05-19-2023-0245 - uniform norm, triangle inequality, discontinuous linear map, weak topology , coherent topology, dual system, dual pair, Transpose of a linear map with respect to pairings, Metrizability and separability, polarity, embedding, equicontinuous, subset, image, adjoint, canonical duality, Space of finite sequences, Reductive dual pair, Role of the axiom of choice, impact, Discontinuous_linear_map, constructivism, A linear map from a finite-dimensional space is always continuous, Classical Invariant Theory, general linear group, centralizer, subgroup, representation theory, isometry group, symplectic vector space, Bounded subsets , strongest polar, bounded convergence, weak convergence, compact subsets, balanced, injective, absolute polar, disk, ring, complex conjugate vector space, coordinates, bilinear, annihilator, orthogonal complement, analogously, degeneracy, convex, initial topology, weakly continuous, etc. (draft)

In topology and related areas of mathematics, the set of all possible topologies on a given set forms a partially ordered set. This order relation can be used for comparison of the topologies. 

https://en.wikipedia.org/wiki/Comparison_of_topologies

From Wikipedia, the free encyclopedia

(Redirected from Weak topology (polar topology))

This article is about the weak topology on a normed vector space. For the weak topology induced by a general family of maps, see initial topology. For the weak topology generated by a cover of a space, see coherent topology.

In mathematics, weak topology is an alternative term for certain initial topologies, often on topological vector spaces or spaces of linear operators, for instance on a Hilbert space. The term is most commonly used for the initial topology of a topological vector space (such as a normed vector space) with respect to its continuous dual. The remainder of this article will deal with this case, which is one of the concepts of functional analysis.

One may call subsets of a topological vector space weakly closed (respectively, weakly compact, etc.) if they are closed (respectively, compact, etc.) with respect to the weak topology. Likewise, functions are sometimes called weakly continuous (respectively, weakly differentiable, weakly analytic, etc.) if they are continuous (respectively, differentiable, analytic, etc.) with respect to the weak topology. 

https://en.wikipedia.org/wiki/Weak_topology

From Wikipedia, the free encyclopedia

This article is about dual pairs of vector spaces. For dual pairs in representation theory, see Reductive dual pair.

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In mathematics, a dual system, dual pair, or 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 ${\displaystyle b:X\times Y\to \mathbb{K}}$. Duality theory, the study of dual systems, is part of functional analysis.

According to Helmut H. Schaefer, "the study of a locally convex space in terms of its dual is the central part of the modern theory of topological vector spaces, for it provides the deepest and most beautiful results of the subject."^[1]

Definition, notation, and conventions

Pairings

A pairing or pair over a field ${\displaystyle \mathbb{K}}$ is a triple ${\displaystyle (X,Y,b),}$ which may also be denoted by ${\displaystyle b(X,Y),}$ consisting of two vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ over ${\displaystyle \mathbb{K}}$ (which this article assumes is either the real numbers ${\displaystyle \mathbb{R}}$ or the complex numbers ${\displaystyle \mathbb{C}}$) and a bilinear map ${\displaystyle b:X\times Y\to \mathbb{K},}$ which is called the bilinear map associated with the pairing^[2] or simply the pairing's map/bilinear form.

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

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

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

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

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

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

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

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

Dual pairings

A pairing ${\displaystyle (X,Y,b)}$ is called a dual system, a dual pair,^[3] 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 ${\displaystyle b(x,y)\ne 0}$);
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 ${\displaystyle b(x,y)\ne 0}$).

In this case say that ${\displaystyle b}$ is non-degenerate, say that ${\displaystyle b}$ places ${\displaystyle X}$ and ${\displaystyle Y}$ in duality (or in separated duality), and ${\displaystyle b}$ is called the duality pairing of the ${\displaystyle (X,Y,b).}$^[2][3]

Total subsets

A subset ${\displaystyle S}$ of ${\displaystyle Y}$ is called total 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]}

Orthogonality

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

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

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

Polar sets

Main article: Polar set

Throughout, ${\displaystyle (X,Y,b)}$ will be a pairing over ${\displaystyle \mathbb{K}.}$ The absolute polar or polar of a subset ${\displaystyle A}$ of ${\displaystyle X}$ is the set:^[4]

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

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

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

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

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 ${\displaystyle Y.}$^[5]

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

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 ${\displaystyle A.}$

Dual definitions and results

Given a pairing ${\displaystyle (X,Y,b),}$ define a new pairing ${\displaystyle (Y,X,d)}$ where ${\displaystyle d(y,x):=b(x,y)}$ for all ${\displaystyle x\in X\text{ and }y\in Y.}$^[2]

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

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

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

For instance, if "${\displaystyle X}$ 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 it allows us to avoid having to assign 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 ${\displaystyle Y}$ (e.g. 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 ${\displaystyle (Y,X,d)}$ (e.g. ${\displaystyle \tau (Y,X,b)}$ actually denotes the topology ${\displaystyle \tau (Y,X,d)}$).

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

Identification of ${\displaystyle (X,Y)}$ with ${\displaystyle (Y,X)}$

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

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 restrictions 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 ${\displaystyle x=0}$).^[2]

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.^[3]

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^[2] — 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.^[2]

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^{[note 2]} 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.^[3]
• 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

Main articles: Weak topology and 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 all maps ${\displaystyle b(\cdot ,y):X\to \mathbb{K}}$ continuous as ${\displaystyle y}$ ranges over ${\displaystyle S.}$^[2] 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, ${\displaystyle {\mathrm{cl}}_{\sigma (X,Y,b)}(S),}$ 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 will arise. 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 ${\displaystyle Y.}$^[2] 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 ${\displaystyle (X,\sigma (X,Y,b)).}$^[2] 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).^[2]

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 ${\displaystyle \sigma (X,Y,b).}$^[2]

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 ${\displaystyle X}$;^[2]
3. ${\displaystyle \sigma (Y,X,b)}$ is Hausdorff.^[2]

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^[2] — 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;^[2]
• ${\displaystyle S\subseteq S^{\perp \perp }=\left({\mathrm{cl}}_{\sigma (X,Y,b)}\mathrm{span}S\right)}$;^[2]
  • 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
  $${\displaystyle {\left(\bigcap _{i\in I}{S}_i\right)}^{\perp }={\mathrm{cl}}_{\sigma (Y,X,b)}\left(\mathrm{span}\left(\bigcup _{i\in I}{S}_i^{\perp }\right)\right).}$$

  
  ^[2]
• If ${\displaystyle {\left(S_i\right)}_{i\in I}}$ is a family of subsets of ${\displaystyle X}$ then ${\displaystyle {\left(\bigcup _{i\in I}{S}_i\right)}^{\perp }=\bigcap _{i\in I}{S}_i^{\perp }.}$^[2]

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 ${\displaystyle X.}$^[2]

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 ${\displaystyle (X,\sigma (X,Y,b)).}$^[6] Furthermore, if ${\displaystyle (X,\sigma (X,Y,b))}$ is a dual system then so is ${\displaystyle \left(M,Y/M^{\perp },b{|}_M\right).}$^[6]

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 ${\displaystyle X/M.}$^[6]

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 ${\displaystyle X.}$^[2]

The following results are important for defining polar topologies.

If ${\displaystyle (X,Y,b)}$ is a pairing and ${\displaystyle A\subseteq X,}$ then:^[2]

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."^[5]
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 ${\displaystyle Y.}$^[7]

Transposes

Transpose of a linear map with respect to pairings

See also: Transpose of a linear map, Transpose, and Transpose § Transposes of linear maps and bilinear forms

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 to 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 ${\displaystyle Z\to Y,}$^{[note 4]} ${\displaystyle X\to Z,}$^{[note 5]} ${\displaystyle W\to Y,}$^{[note 6]} ${\displaystyle Y\to W,}$^{[note 7]} etc. (see footnote for details).

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 ${\displaystyle \left(W,\sigma \left(W,Z,c\right)\right).}$^[2]
• 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 ${\displaystyle {}^t\left(F^{-1}\right)={\left({}^{t}F\right)}^{-1}}$^[2]
• Let ${\displaystyle S\subseteq X}$ and let ${\displaystyle S^{\circ }}$ denotes the absolute polar of ${\displaystyle A,}$ then:^[2]
  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 ${\displaystyle \Vert F\Vert =\Vert {}^{t}F\Vert .}$^[2]

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.^[2]

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).^[2]

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.^[2] 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.^[2]

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.^[3]

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 ${\displaystyle w^{\mathrm{\prime }}\in W^{\mathrm{\#}},}$ ${\displaystyle F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)=w^{\mathrm{\prime }}\circ F}$^[2][8] 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, ${\displaystyle F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)(x)=w^{\mathrm{\prime }}(F(x))\text{ for all }x\in X.}$

Examples

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:^[2]

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.

If ${\displaystyle F}$ is weakly continuous then ${\displaystyle {}^{t}F::(Z,\sigma (Z,W))\to (Y,\sigma (Y,X))}$ will be continuous and furthermore, ${\displaystyle {}^{tt}F=F}$^[8]

A map ${\displaystyle g:A\to B}$ between topological spaces is relatively open if ${\displaystyle g:A\to \mathrm{Im}g}$ is an open mapping, where ${\displaystyle \mathrm{Im}g}$ is the range of ${\displaystyle g.}$^[2]

Suppose that ${\displaystyle ⟨X,Y⟩}$ and ${\displaystyle ⟨W,Z⟩}$ are dual systems and ${\displaystyle F:X\to W}$ is a weakly continuous linear map. Then the following are equivalent:^[2]

1. ${\displaystyle F:(X,\sigma (X,Y))\to (W,\sigma (W,Z))}$ is relatively open;
2. The range of ${\displaystyle {}^{t}F}$ is ${\displaystyle \sigma (Y,X)}$-closed in ${\displaystyle Y}$;
3. ${\displaystyle \mathrm{Im}{}^{t}F=(\ker F{)}^{\perp }}$

Furthermore,

• ${\displaystyle F:X\to W}$ is injective (resp. bijective) if and only if ${\displaystyle {}^{t}F}$ is surjective (resp. bijective);
• ${\displaystyle F:X\to W}$ is surjective if and only if ${\displaystyle {}^{t}F::(Z,\sigma (Z,W))\to (Y,\sigma (Y,X))}$ is relatively open and injective.

Transpose of a map between TVSs

The transpose of map between two TVSs is defined if and only if ${\displaystyle F}$ is weakly continuous.

If ${\displaystyle F:X\to Y}$ is a linear map between two Hausdorff locally convex topological vector spaces then:^[2]

• If ${\displaystyle F}$ is continuous then it is weakly continuous and ${\displaystyle {}^{t}F}$ is both Mackey continuous and strongly continuous.
• If ${\displaystyle F}$ is weakly continuous then it is both Mackey continuous and strongly continuous (defined below).
• If ${\displaystyle F}$ is weakly continuous then it is continuous if and only if ${\displaystyle {}^{t}F{:}^{\mathrm{\prime }}\to X^{\mathrm{\prime }}}$ maps equicontinuous subsets of ${\displaystyle Y^{\mathrm{\prime }}}$ to equicontinuous subsets of ${\displaystyle X^{\mathrm{\prime }}.}$
• If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces then ${\displaystyle F}$ is continuous if and only if it is weakly continuous, in which case ${\displaystyle \Vert F\Vert =\Vert {}^{t}F\Vert .}$
• If ${\displaystyle F}$ is continuous then ${\displaystyle F:X\to Y}$ is relatively open if and only if ${\displaystyle F}$ is weakly relatively open (i.e. ${\displaystyle F:\left(X,\sigma \left(X,X^{\mathrm{\prime }}\right)\right)\to \left(Y,\sigma \left(Y,Y^{\mathrm{\prime }}\right)\right)}$ is relatively open) and every equicontinuous subsets of ${\displaystyle \mathrm{Im}{}^{t}F={}^{t}F\left(Y^{\mathrm{\prime }}\right)}$ is the image of some equicontinuous subsets of ${\displaystyle Y^{\mathrm{\prime }}.}$
• If ${\displaystyle F}$ is continuous injection then ${\displaystyle F:X\to Y}$ is a TVS-embedding (or equivalently, a topological embedding) if and only if every equicontinuous subsets of ${\displaystyle X^{\mathrm{\prime }}}$ is the image of some equicontinuous subsets of ${\displaystyle Y^{\mathrm{\prime }}.}$

Metrizability and separability

Let ${\displaystyle X}$ be a locally convex space with continuous dual space ${\displaystyle X^{\mathrm{\prime }}}$ and let ${\displaystyle K\subseteq X^{\mathrm{\prime }}.}$^[2]

1. If ${\displaystyle K}$ is equicontinuous or ${\displaystyle \sigma \left(X^{\mathrm{\prime }},X\right)}$-compact, and if ${\displaystyle D\subseteq X^{\mathrm{\prime }}}$ is such that ${\displaystyle \mathrm{span}D}$is dense in ${\displaystyle X,}$ then the subspace topology that ${\displaystyle K}$ inherits from ${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},D\right)\right)}$ is identical to the subspace topology that ${\displaystyle K}$ inherits from ${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right).}$
2. If ${\displaystyle X}$ is separable and ${\displaystyle K}$ is equicontinuous then ${\displaystyle K,}$ when endowed with the subspace topology induced by ${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right),}$ is metrizable.
3. If ${\displaystyle X}$ is separable and metrizable, then ${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right)}$ is separable.
4. If ${\displaystyle X}$ is a normed space then ${\displaystyle X}$ is separable if and only if the closed unit call the continuous dual space of ${\displaystyle X}$ is metrizable when given the subspace topology induced by ${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right).}$
5. If ${\displaystyle X}$ is a normed space whose continuous dual space is separable (when given the usual norm topology), then ${\displaystyle X}$ is separable.

Polar topologies and topologies compatible with pairing

Main article: Polar topology

Starting with only the weak topology, the use of polar sets produces a range of locally convex topologies. Such topologies are called polar topologies. The weak topology is the weakest topology of this range.

Throughout, ${\displaystyle (X,Y,b)}$ will be a pairing over ${\displaystyle \mathbb{K}}$ and ${\displaystyle \mathcal{G}}$ will be a non-empty collection of ${\displaystyle \sigma (X,Y,b)}$-bounded subsets of ${\displaystyle X.}$

Polar topologies

Main article: Polar topology

Given a collection ${\displaystyle \mathcal{G}}$ of subsets of ${\displaystyle X}$, the polar topology on ${\displaystyle Y}$ determined by ${\displaystyle \mathcal{G}}$ (and ${\displaystyle b}$) or the ${\displaystyle \mathcal{G}}$-topology on ${\displaystyle Y}$ is the unique topological vector space (TVS) topology on ${\displaystyle Y}$ for which

$${\displaystyle \left\{r{G}^{\circ }:G\in \mathcal{G},r>0\right\}}$$

forms a subbasis of neighborhoods at the origin.^[2] When ${\displaystyle Y}$ is endowed with this ${\displaystyle \mathcal{G}}$-topology then it is denoted by Y_{${\displaystyle \mathcal{G}}$}. Every polar topology is necessarily locally convex.^[2] When ${\displaystyle \mathcal{G}}$ is a directed set with respect to subset inclusion (i.e. if for all ${\displaystyle G,K\in \mathcal{G}}$ there exists some ${\displaystyle K\in \mathcal{G}}$ such that ${\displaystyle G\cup H\subseteq K}$) then this neighborhood subbasis at 0 actually forms a neighborhood basis at 0.^[2]

The following table lists some of the more important polar topologies.

Notation: If ${\displaystyle \mathrm{\Delta }(X,Y,b)}$ denotes a polar topology on ${\displaystyle Y}$ then ${\displaystyle Y}$ endowed with this topology will be denoted by ${\displaystyle Y_{\mathrm{\Delta }(Y,X,b)},}$ ${\displaystyle Y_{\mathrm{\Delta }(Y,X)}}$ or simply ${\displaystyle Y_{\mathrm{\Delta }}}$ (e.g. for ${\displaystyle \sigma (Y,X,b)}$ we'd have ${\displaystyle \mathrm{\Delta }=\sigma }$ so that ${\displaystyle Y_{\sigma (Y,X,b)},}$ ${\displaystyle Y_{\sigma (Y,X)}}$ and ${\displaystyle Y_{\sigma }}$ all denote ${\displaystyle Y}$ endowed with ${\displaystyle \sigma (X,Y,b)}$).

${\displaystyle \mathcal{G}\subseteq \mathcal{P}X}$ ("topology of uniform convergence on ...")	Notation	Name ("topology of...")	Alternative name
finite subsets of ${\displaystyle X}$ (or ${\displaystyle \sigma (X,Y,b)}$-closed disked hulls of finite subsets of ${\displaystyle X}$)	${\displaystyle \sigma (X,Y,b)}$ ${\displaystyle s(X,Y,b)}$	pointwise/simple convergence	weak/weak* topology
${\displaystyle \sigma (X,Y,b)}$-compact disks	${\displaystyle \tau (X,Y,b)}$		Mackey topology
${\displaystyle \sigma (X,Y,b)}$-compact convex subsets	${\displaystyle \gamma (X,Y,b)}$	compact convex convergence
${\displaystyle \sigma (X,Y,b)}$-compact subsets (or balanced ${\displaystyle \sigma (X,Y,b)}$-compact subsets)	${\displaystyle c(X,Y,b)}$	compact convergence
${\displaystyle \sigma (X,Y,b)}$-bounded subsets	${\displaystyle b(X,Y,b)}$ ${\displaystyle \beta (X,Y,b)}$	bounded convergence	strong topology Strongest polar topology

Definitions involving polar topologies

Continuity

A linear map ${\displaystyle F:X\to W}$ is Mackey continuous (with respect to ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$) if ${\displaystyle F:(X,\tau (X,Y,b))\to (W,\tau (W,Z,c))}$ is continuous.^[2]

A linear map ${\displaystyle F:X\to W}$ is strongly continuous (with respect to ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$) if ${\displaystyle F:(X,\beta (X,Y,b))\to (W,\beta (W,Z,c))}$ is continuous.^[2]

Bounded subsets

A subset of ${\displaystyle X}$ is weakly bounded (resp. Mackey bounded, strongly bounded) if it is bounded in ${\displaystyle (X,\sigma (X,Y,b))}$ (resp. bounded in ${\displaystyle (X,\tau (X,Y,b)),}$ bounded in ${\displaystyle (X,\beta (X,Y,b))}$).

Topologies compatible with a pair

If ${\displaystyle (X,Y,b)}$ is a pairing over ${\displaystyle \mathbb{K}}$ and ${\displaystyle \mathcal{T}}$ is a vector topology on ${\displaystyle X}$ then ${\displaystyle \mathcal{T}}$ is a topology of the pairing and that it is compatible (or consistent) with the pairing ${\displaystyle (X,Y,b)}$ if it is locally convex and if the continuous dual space of ${\displaystyle \left(X,\mathcal{T}\right)=b(\cdot ,Y).}$^{[note 8]} If ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$ then by identifying ${\displaystyle Y}$ as a vector subspace of ${\displaystyle X}$'s algebraic dual, the defining condition becomes: ${\displaystyle {\left(X,\mathcal{T}\right)}^{\mathrm{\prime }}=Y.}$^[2] Some authors (e.g. [Trèves 2006] and [Schaefer 1999]) require that a topology of a pair also be Hausdorff,^[3][9] which it would have to be if ${\displaystyle Y}$ distinguishes the points of ${\displaystyle X}$ (which these authors assume).

The weak topology ${\displaystyle \sigma (X,Y,b)}$ is compatible with the pairing ${\displaystyle (X,Y,b)}$ (as was shown in the Weak representation theorem) and it is in fact the weakest such topology. There is a strongest topology compatible with this pairing and that is the Mackey topology. If ${\displaystyle N}$ is a normed space that is not reflexive then the usual norm topology on its continuous dual space is not compatible with the duality ${\displaystyle \left(N^{\mathrm{\prime }},N\right).}$^[2]

Mackey–Arens theorem

Main articles: Mackey–Arens theorem, Mackey topology, and Mackey space

The following is one of the most important theorems in duality theory.

Mackey–Arens theorem I^[2] — Let ${\displaystyle (X,Y,b)}$ will be a pairing such that ${\displaystyle X}$ distinguishes the points of ${\displaystyle Y}$ and let ${\displaystyle \mathcal{T}}$ be a locally convex topology on ${\displaystyle X}$ (not necessarily Hausdorff). Then ${\displaystyle \mathcal{T}}$ is compatible with the pairing ${\displaystyle (X,Y,b)}$ if and only if ${\displaystyle \mathcal{T}}$ is a polar topology determined by some collection ${\displaystyle \mathcal{G}}$ of ${\displaystyle \sigma (Y,X,b)}$-compact disks that cover^{[note 9]} ${\displaystyle Y.}$

It follows that the Mackey topology ${\displaystyle \tau (X,Y,b),}$ which recall is the polar topology generated by all ${\displaystyle \sigma (X,Y,b)}$-compact disks in ${\displaystyle Y,}$ is the strongest locally convex topology on ${\displaystyle X}$ that is compatible with the pairing ${\displaystyle (X,Y,b).}$ A locally convex space whose given topology is identical to the Mackey topology is called a Mackey space. The following consequence of the above Mackey-Arens theorem is also called the Mackey-Arens theorem.

Mackey–Arens theorem II^[2] — Let ${\displaystyle (X,Y,b)}$ will be a pairing such that ${\displaystyle X}$ distinguishes the points of ${\displaystyle Y}$ and let ${\displaystyle \mathcal{T}}$ be a locally convex topology on ${\displaystyle X.}$ Then ${\displaystyle \mathcal{T}}$ is compatible with the pairing if and only if ${\displaystyle \sigma (X,Y,b)\subseteq \mathcal{T}\subseteq \tau (X,Y,b).}$

Mackey's theorem, barrels, and closed convex sets

If ${\displaystyle X}$ is a TVS (over ${\displaystyle \mathbb{R}}$ or ${\displaystyle \mathbb{C}}$) then a half-space is a set of the form ${\displaystyle \{x\in X:f(x)\le r\}}$ for some real ${\displaystyle r}$ and some continuous real linear functional ${\displaystyle f}$ on ${\displaystyle X.}$

Theorem — If ${\displaystyle X}$ is a locally convex space (over ${\displaystyle \mathbb{R}}$ or ${\displaystyle \mathbb{C}}$) and if ${\displaystyle C}$ is a non-empty closed and convex subset of ${\displaystyle X,}$ then ${\displaystyle C}$ is equal to the intersection of all closed half spaces containing it.^[10]

The above theorem implies that the closed and convex subsets of a locally convex space depend entirely on the continuous dual space. Consequently, the closed and convex subsets are the same in any topology compatible with duality; that is, if ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{L}}$ are any locally convex topologies on ${\displaystyle X}$ with the same continuous dual spaces, then a convex subset of ${\displaystyle X}$ is closed in the ${\displaystyle \mathcal{T}}$ topology if and only if it is closed in the ${\displaystyle \mathcal{L}}$ topology. This implies that the ${\displaystyle \mathcal{T}}$-closure of any convex subset of ${\displaystyle X}$ is equal to its ${\displaystyle \mathcal{L}}$-closure and that for any ${\displaystyle \mathcal{T}}$-closed disk ${\displaystyle A}$ in ${\displaystyle X,}$ ${\displaystyle A=A^{\circ \circ }.}$^[2] In particular, if ${\displaystyle B}$ is a subset of ${\displaystyle X}$ then ${\displaystyle B}$ is a barrel in ${\displaystyle (X,\mathcal{L})}$ if and only if it is a barrel in ${\displaystyle (X,\mathcal{L}).}$^[2]

The following theorem shows that barrels (i.e. closed absorbing disks) are exactly the polars of weakly bounded subsets.

Theorem^[2] — Let ${\displaystyle (X,Y,b)}$ will be a pairing such that ${\displaystyle X}$ distinguishes the points of ${\displaystyle Y}$ and let ${\displaystyle \mathcal{T}}$ be a topology of the pair. Then a subset of ${\displaystyle X}$ is a barrel in ${\displaystyle X}$ if and only if it is equal to the polar of some ${\displaystyle \sigma (Y,X,b)}$-bounded subset of ${\displaystyle Y.}$

If ${\displaystyle X}$ is a topological vector space then:^[2][11]

1. A closed absorbing and balanced subset ${\displaystyle B}$ of ${\displaystyle X}$ absorbs each convex compact subset of ${\displaystyle X}$ (i.e. there exists a real ${\displaystyle r>0}$ such that ${\displaystyle rB}$ contains that set).
2. If ${\displaystyle X}$ is Hausdorff and locally convex then every barrel in ${\displaystyle X}$ absorbs every convex bounded complete subset of ${\displaystyle X.}$

All of this leads to Mackey's theorem, which is one of the central theorems in the theory of dual systems. In short, it states the bounded subsets are the same for any two Hausdorff locally convex topologies that are compatible with the same duality.

Mackey's theorem^[11][2] — Suppose that ${\displaystyle (X,\mathcal{L})}$ is a Hausdorff locally convex space with continuous dual space ${\displaystyle X^{\mathrm{\prime }}}$ and consider the canonical duality ${\displaystyle ⟨X,X^{\mathrm{\prime }}⟩.}$ If ${\displaystyle \mathcal{L}}$ is any topology on ${\displaystyle X}$ that is compatible with the duality ${\displaystyle ⟨X,X^{\mathrm{\prime }}⟩}$ on ${\displaystyle X}$ then the bounded subsets of ${\displaystyle (X,\mathcal{L})}$ are the same as the bounded subsets of ${\displaystyle (X,\mathcal{L}).}$

Examples

Space of finite sequences

Let ${\displaystyle X}$ denote the space of all sequences of scalars ${\displaystyle r_{\bullet }={\left(r_i\right)}_{i=1}^{\mathrm{\infty }}}$ such that ${\displaystyle r_i=0}$ for all sufficiently large ${\displaystyle i.}$ Let ${\displaystyle Y=X}$ and define a bilinear map ${\displaystyle b:X\times X\to \mathbb{K}}$ by

$${\displaystyle b\left(r_{\bullet },s_{\bullet }\right):=\sum _{i=1}^{\mathrm{\infty }}{r}_i{s}_i.}$$

Then ${\displaystyle \sigma (X,X,b)=\tau (X,X,b).}$^[2] Moreover, a subset ${\displaystyle T\subseteq X}$ is ${\displaystyle \sigma (X,X,b)}$-bounded (resp. ${\displaystyle \beta (X,X,b)}$-bounded) if and only if there exists a sequence ${\displaystyle m_{\bullet }={\left(m_i\right)}_{i=1}^{\mathrm{\infty }}}$ of positive real numbers such that ${\displaystyle \left|t_i\right|\le m_i}$ for all ${\displaystyle t_{\bullet }={\left(t_i\right)}_{i=1}^{\mathrm{\infty }}\in T}$ and all indices ${\displaystyle i}$ (resp. and ${\displaystyle m_{\bullet }\in X}$).^[2] It follows that there are weakly bounded (that is, ${\displaystyle \sigma (X,X,b)}$-bounded) subsets of ${\displaystyle X}$ that are not strongly bounded (that is, not ${\displaystyle \beta (X,X,b)}$-bounded).

See also

• Biorthogonal system
• Dual space – In mathematics, vector space of linear forms
• Dual topology
• Duality (mathematics) – General concept and operation in mathematics
• Inner product – Generalization of the dot product; used to define Hilbert spaces
• L-semi-inner product – Generalization of inner products that applies to all normed spaces
• Pairing – bilinear map of modules over a ring
• Polar set – Subset of all points that is bounded by some given point of a dual (in a dual pairing)
• Polar topology – Dual space topology of uniform convergence on some sub-collection of bounded subsets
• Reductive dual pair
• Strong dual space – Continuous dual space endowed with the topology of uniform convergence on bounded sets
• Strong topology (polar topology) – Continuous dual space endowed with the topology of uniform convergence on bounded sets
• Topologies on spaces of linear maps
• Weak topology – Mathematical term

Notes

1. 
   

A subset ${\displaystyle S}$ of ${\displaystyle X}$ is total if for all ${\displaystyle y\in Y,}$

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

implies ${\displaystyle y=0.}$

That ${\displaystyle b}$ is linear in its first coordinate is obvious. Suppose ${\displaystyle c}$ is a scalar. Then ${\displaystyle b(x,c\perp y)=b\left(x,\overline{c}y\right)=⟨x,\overline{c}y⟩=c⟨x,y⟩=cb(x,y),}$ which shows that ${\displaystyle b}$ is linear in its second coordinate.

The weak topology on ${\displaystyle Y}$ is the weakest TVS topology on ${\displaystyle Y}$ making all maps ${\displaystyle b(x,\cdot ):Y\to \mathbb{K}}$ continuous, as ${\displaystyle x}$ ranges over ${\displaystyle R.}$ The dual notation of ${\displaystyle (Y,\sigma (Y,R,b)),}$ ${\displaystyle (Y,\sigma (Y,R)),}$ or simply ${\displaystyle (Y,\sigma )}$ may also be used to denote ${\displaystyle Y}$ endowed with the weak topology ${\displaystyle \sigma (Y,R,b).}$ If ${\displaystyle R}$ is not clear from context then it should be assumed to be all of ${\displaystyle X,}$ in which case it is simply called the weak topology on ${\displaystyle Y}$ (induced by ${\displaystyle X}$).

If ${\displaystyle G:Z\to Y}$ is a linear map then ${\displaystyle G}$'s transpose, ${\displaystyle {}^{t}G:X\to W,}$ is well-defined if and only if ${\displaystyle Z}$ distinguishes points of ${\displaystyle W}$ and ${\displaystyle b(X,G(\cdot ))\subseteq c(W,\cdot ).}$ In this case, for each ${\displaystyle x\in X,}$ the defining condition for ${\displaystyle {}^{t}G(x)}$ is: ${\displaystyle c(x,G(\cdot ))=c\left({}^{t}G(x),\cdot \right).}$

If ${\displaystyle H:X\to Z}$ is a linear map then ${\displaystyle H}$'s transpose, ${\displaystyle {}^{t}H:W\to Y,}$ is well-defined if and only if ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$ and ${\displaystyle c(W,H(\cdot ))\subseteq b(\cdot ,Y).}$ In this case, for each ${\displaystyle w\in W,}$ the defining condition for ${\displaystyle {}^{t}H(w)}$ is: ${\displaystyle c(w,H(\cdot ))=b\left(\cdot ,{}^{t}H(w)\right).}$

If ${\displaystyle H:W\to Y}$ is a linear map then ${\displaystyle H}$'s transpose, ${\displaystyle {}^{t}H:X\to Q,}$ is well-defined if and only if ${\displaystyle W}$ distinguishes points of ${\displaystyle Z}$ and ${\displaystyle b(X,H(\cdot ))\subseteq c(\cdot ,Z).}$ In this case, for each ${\displaystyle x\in X,}$ the defining condition for ${\displaystyle {}^{t}H(x)}$ is: ${\displaystyle c(x,H(\cdot ))=b\left(\cdot ,{}^{t}H(x)\right).}$

If ${\displaystyle H:Y\to W}$ is a linear map then ${\displaystyle H}$'s transpose, ${\displaystyle {}^{t}H:Z\to X,}$ is well-defined if and only if ${\displaystyle Y}$ distinguishes points of ${\displaystyle X}$ and ${\displaystyle c(H(\cdot ),Z)\subseteq b(X,\cdot ).}$ In this case, for each ${\displaystyle z\in Z,}$ the defining condition for ${\displaystyle {}^{t}H(z)}$ is: ${\displaystyle c(H(\cdot ),z)=b\left({}^{t}H(z),\cdot \right).}$

Of course, there is an analogous definition for topologies on ${\displaystyle Y}$ to be "compatible it a pairing" but this article will only deal with topologies on ${\displaystyle X.}$

9. Recall that a collection of subsets of a set ${\displaystyle S}$ is said to cover ${\displaystyle S}$ if every point of ${\displaystyle S}$ is contained in some set belonging to the collection.

References

1. 
   

Schaefer & Wolff 1999, p. 122.

Narici & Beckenstein 2011, pp. 225–273.

Schaefer & Wolff 1999, pp. 122–128.

Trèves 2006, p. 195.

Schaefer & Wolff 1999, pp. 123–128.

Narici & Beckenstein 2011, pp. 260–264.

Narici & Beckenstein 2011, pp. 251–253.

Schaefer & Wolff 1999, pp. 128–130.

Trèves 2006, pp. 368–377.

Narici & Beckenstein 2011, p. 200.

11. Trèves 2006, pp. 371–372.

Bibliography

• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, Vol. 1, Functional Analysis, Section III.3. Academic Press, San Diego, 1980. ISBN 0-12-585050-6.
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Schmitt, Lothar M (1992). "An Equivariant Version of the Hahn–Banach Theorem". Houston J. Of Math. 18: 429–447.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.

External links

• Duality Theory

v t e Functional analysis (topics – glossary)

v t e Convex analysis and variational analysis

v t e Duality and spaces of linear maps

v t e Topological vector spaces (TVSs)

Categories:

• Functional analysis
• Duality theories
• Linear functionals
• Topological vector spaces

 

 https://en.wikipedia.org/wiki/Dual_system#Weak_continuity

In the mathematical field of representation theory, a reductive dual pair is a pair of subgroups (G, G′) of the isometry group Sp(W) of a symplectic vector space W, such that G is the centralizer of G′ in Sp(W) and vice versa, and these groups act reductively on W. Somewhat more loosely, one speaks of a dual pair whenever two groups are the mutual centralizers in a larger group, which is frequently a general linear group. The concept was introduced by Roger Howe in Howe (1979). Its strong ties with Classical Invariant Theory are discussed in Howe (1989a). 

https://en.wikipedia.org/wiki/Reductive_dual_pair

In Weyl's wonderful and terrible¹ book The Classical Groups [W] one may discern two main themes: first, the study of the polynomial invariants for an arbitrary number of (contravariant or covariant) variables for a standard classical group action; second, the isotypic decomposition of the full tensor algebra for such an action.

¹Most people who know the book feel the material in it is wonderful. Many also feel the presentation is terrible. (The author is not among these latter.)

Howe (1989, p.539)

The Classical Groups: Their Invariants and Representations is a mathematics book by Hermann Weyl (1939), which describes classical invariant theory in terms of representation theory. It is largely responsible for the revival of interest in invariant theory, which had been almost killed off by David Hilbert's solution of its main problems in the 1890s.

Weyl (1939a) gave an informal talk about the topic of his book. There was a second edition in 1946. 

https://en.wikipedia.org/wiki/The_Classical_Groups

Representation theory is a branch of mathematics that studies abstract algebraic structures by representing their elements as linear transformations of vector spaces, and studies modules over these abstract algebraic structures.^[1][2] In essence, a representation makes an abstract algebraic object more concrete by describing its elements by matrices and their algebraic operations (for example, matrix addition, matrix multiplication). The theory of matrices and linear operators is well-understood, so representations of more abstract objects in terms of familiar linear algebra objects helps glean properties and sometimes simplify calculations on more abstract theories.  

https://en.wikipedia.org/wiki/Representation_theory

In mathematics, specifically linear algebra, a degenerate bilinear form f (x, y ) on a vector space V is a bilinear form such that the map from V to V^∗ (the dual space of V ) given by v ↦ (x ↦ f (x, v )) is not an isomorphism. An equivalent definition when V is finite-dimensional is that it has a non-trivial kernel: there exist some non-zero x in V such that

${\displaystyle f(x,y)=0}$ for all ${\displaystyle y\in V.}$

 https://en.wikipedia.org/wiki/Degenerate_bilinear_form

https://en.wikipedia.org/wiki/Strong_dual_space

https://en.wikipedia.org/wiki/Normed_vector_space

https://en.wikipedia.org/wiki/Coherent_topology

https://en.wikipedia.org/wiki/Initial_topology

https://en.wikipedia.org/wiki/Dual_space#Continuous_dual_space

https://en.wikipedia.org/wiki/Dual_space#Continuous_dual_space

 

 https://en.wikipedia.org/wiki/Discontinuous_linear_map

 

In mathematics, linear maps form an important class of "simple" functions which preserve the algebraic structure of linear spaces and are often used as approximations to more general functions (see linear approximation). If the spaces involved are also topological spaces (that is, topological vector spaces), then it makes sense to ask whether all linear maps are continuous. It turns out that for maps defined on infinite-dimensional topological vector spaces (e.g., infinite-dimensional normed spaces), the answer is generally no: there exist discontinuous linear maps. If the domain of definition is complete, it is trickier; such maps can be proven to exist, but the proof relies on the axiom of choice and does not provide an explicit example.

A linear map from a finite-dimensional space is always continuous

Let X and Y be two normed spaces and ${\displaystyle f:X\to Y}$ a linear map from X to Y. If X is finite-dimensional, choose a basis ${\displaystyle \left(e_1,e_2,\dots ,e_n\right)}$ in X which may be taken to be unit vectors. Then,

$${\displaystyle f(x)=\sum _{i=1}^n{x}_{i}f(e_i),}$$

and so by the triangle inequality,

$${\displaystyle \Vert f(x)\Vert =\Vert \sum _{i=1}^n{x}_{i}f(e_i)\Vert \le \sum _{i=1}^n|x_i|\Vert f(e_i)\Vert .}$$

Letting

$${\displaystyle M=\underset{i}{sup}\{\Vert f(e_i)\Vert \},}$$

and using the fact that

$${\displaystyle \sum _{i=1}^n|x_i|\le C\Vert x\Vert }$$

for some C>0 which follows from the fact that any two norms on a finite-dimensional space are equivalent, one finds

$${\displaystyle \Vert f(x)\Vert \le \left(\sum _{i=1}^n|x_i|\right)M\le CM\Vert x\Vert .}$$

Thus, ${\displaystyle f}$ is a bounded linear operator and so is continuous. In fact, to see this, simply note that f is linear, and therefore ${\displaystyle \Vert f(x)-f(x^{\prime })\Vert =\Vert f(x-x^{\prime })\Vert \le K\Vert x-x^{\prime }\Vert }$ for some universal constant K. Thus for any ${\displaystyle ϵ>0,}$ we can choose ${\displaystyle \delta \le ϵ/K}$ so that ${\displaystyle f(B(x,\delta ))\subseteq B(f(x),ϵ)}$ (${\displaystyle B(x,\delta )}$ and ${\displaystyle B(f(x),ϵ)}$ are the normed balls around ${\displaystyle x}$ and ${\displaystyle f(x)}$), which gives continuity.

If X is infinite-dimensional, this proof will fail as there is no guarantee that the supremum M exists. If Y is the zero space {0}, the only map between X and Y is the zero map which is trivially continuous. In all other cases, when X is infinite-dimensional and Y is not the zero space, one can find a discontinuous map from X to Y.

A concrete example

Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence ${\displaystyle e_i}$ of linearly independent vectors which does not have a limit, there is a linear operator ${\displaystyle T}$ such that the quantities ${\displaystyle \Vert T(e_i)\Vert /\Vert e_i\Vert }$ grow without bound. In a sense, the linear operators are not continuous because the space has "holes".

For example, consider the space X of real-valued smooth functions on the interval [0, 1] with the uniform norm, that is,

$${\displaystyle \Vert f\Vert =\underset{x\in [0,1]}{sup}|f(x)|.}$$

The derivative-at-a-point map, given by

$${\displaystyle T(f)=f^{\prime }(0)}$$

defined on X and with real values, is linear, but not continuous. Indeed, consider the sequence

$${\displaystyle f_n(x)=\frac{\sin (n^{2}x)}{n}}$$

for ${\displaystyle n\ge 1.}$ This sequence converges uniformly to the constantly zero function, but

$${\displaystyle T(f_n)=\frac{n^2\cos (n^2\cdot 0)}{n}=n\to \mathrm{\infty }}$$

as ${\displaystyle n\to \mathrm{\infty }}$ instead of ${\displaystyle T(f_n)\to T(0)=0}$ which would hold for a continuous map. Note that T is real-valued, and so is actually a linear functional on X (an element of the algebraic dual space X^*). The linear map X → X which assigns to each function its derivative is similarly discontinuous. Note that although the derivative operator is not continuous, it is closed.

The fact that the domain is not complete here is important. Discontinuous operators on complete spaces require a little more work.

A nonconstructive example

An algebraic basis for the real numbers as a vector space over the rationals is known as a Hamel basis (note that some authors use this term in a broader sense to mean an algebraic basis of any vector space). Note that any two noncommensurable numbers, say 1 and ${\displaystyle \pi }$, are linearly independent. One may find a Hamel basis containing them, and define a map ${\displaystyle f:\mathbb{R}\to R}$ so that ${\displaystyle f(\pi )=0,}$ f acts as the identity on the rest of the Hamel basis, and extend to all of ${\displaystyle \mathbb{R}}$ by linearity. Let {r_n}_n be any sequence of rationals which converges to ${\displaystyle \pi }$. Then lim_n f(r_n) = π, but ${\displaystyle f(\pi )=0.}$ By construction, f is linear over ${\displaystyle \mathbb{Q}}$ (not over ${\displaystyle \mathbb{R}}$), but not continuous. Note that f is also not measurable; an additive real function is linear if and only if it is measurable, so for every such function there is a Vitali set. The construction of f relies on the axiom of choice.

This example can be extended into a general theorem about the existence of discontinuous linear maps on any infinite-dimensional normed space (as long as the codomain is not trivial).

General existence theorem

Discontinuous linear maps can be proven to exist more generally, even if the space is complete. Let X and Y be normed spaces over the field K where ${\displaystyle K=\mathbb{R}}$ or ${\displaystyle K=\mathbb{C}.}$ Assume that X is infinite-dimensional and Y is not the zero space. We will find a discontinuous linear map f from X to K, which will imply the existence of a discontinuous linear map g from X to Y given by the formula ${\displaystyle g(x)=f(x)y_0}$ where ${\displaystyle y_0}$ is an arbitrary nonzero vector in Y.

If X is infinite-dimensional, to show the existence of a linear functional which is not continuous then amounts to constructing f which is not bounded. For that, consider a sequence (e_n)_n (${\displaystyle n\ge 1}$) of linearly independent vectors in X. Define

$${\displaystyle T(e_n)=n\Vert e_n\Vert }$$

for each ${\displaystyle n=1,2,\dots }$ Complete this sequence of linearly independent vectors to a vector space basis of X, and define T at the other vectors in the basis to be zero. T so defined will extend uniquely to a linear map on X, and since it is clearly not bounded, it is not continuous.

Notice that by using the fact that any set of linearly independent vectors can be completed to a basis, we implicitly used the axiom of choice, which was not needed for the concrete example in the previous section but one.

Role of the axiom of choice

As noted above, the axiom of choice (AC) is used in the general existence theorem of discontinuous linear maps. In fact, there are no constructive examples of discontinuous linear maps with complete domain (for example, Banach spaces). In analysis as it is usually practiced by working mathematicians, the axiom of choice is always employed (it is an axiom of ZFC set theory); thus, to the analyst, all infinite-dimensional topological vector spaces admit discontinuous linear maps.

On the other hand, in 1970 Robert M. Solovay exhibited a model of set theory in which every set of reals is measurable.^[1] This implies that there are no discontinuous linear real functions. Clearly AC does not hold in the model.

Solovay's result shows that it is not necessary to assume that all infinite-dimensional vector spaces admit discontinuous linear maps, and there are schools of analysis which adopt a more constructivist viewpoint. For example, H. G. Garnir, in searching for so-called "dream spaces" (topological vector spaces on which every linear map into a normed space is continuous), was led to adopt ZF + DC + BP (dependent choice is a weakened form and the Baire property is a negation of strong AC) as his axioms to prove the Garnir–Wright closed graph theorem which states, among other things, that any linear map from an F-space to a TVS is continuous. Going to the extreme of constructivism, there is Ceitin's theorem, which states that every function is continuous (this is to be understood in the terminology of constructivism, according to which only representable functions are considered to be functions).^[2] Such stances are held by only a small minority of working mathematicians.

The upshot is that the existence of discontinuous linear maps depends on AC; it is consistent with set theory without AC that there are no discontinuous linear maps on complete spaces. In particular, no concrete construction such as the derivative can succeed in defining a discontinuous linear map everywhere on a complete space.

Closed operators

Many naturally occurring linear discontinuous operators are closed, a class of operators which share some of the features of continuous operators. It makes sense to ask which linear operators on a given space are closed. The closed graph theorem asserts that an everywhere-defined closed operator on a complete domain is continuous, so to obtain a discontinuous closed operator, one must permit operators which are not defined everywhere.

To be more concrete, let ${\displaystyle T}$ be a map from ${\displaystyle X}$ to ${\displaystyle Y}$ with domain ${\displaystyle \mathrm{Dom}(T),}$ written ${\displaystyle T:\mathrm{Dom}(T)\subseteq X\to Y.}$ We don't lose much if we replace X by the closure of ${\displaystyle \mathrm{Dom}(T).}$ That is, in studying operators that are not everywhere-defined, one may restrict one's attention to densely defined operators without loss of generality.

If the graph ${\displaystyle \mathrm{\Gamma }(T)}$ of ${\displaystyle T}$ is closed in ${\displaystyle X\times Y,}$ we call T closed. Otherwise, consider its closure ${\displaystyle \overline{\mathrm{\Gamma }(T)}}$ in ${\displaystyle X\times Y.}$ If ${\displaystyle \overline{\mathrm{\Gamma }(T)}}$ is itself the graph of some operator ${\displaystyle \overline{T},}$ ${\displaystyle T}$ is called closable, and ${\displaystyle \overline{T}}$ is called the closure of ${\displaystyle T.}$

So the natural question to ask about linear operators that are not everywhere-defined is whether they are closable. The answer is, "not necessarily"; indeed, every infinite-dimensional normed space admits linear operators that are not closable. As in the case of discontinuous operators considered above, the proof requires the axiom of choice and so is in general nonconstructive, though again, if X is not complete, there are constructible examples.

In fact, there is even an example of a linear operator whose graph has closure all of ${\displaystyle X\times Y.}$ Such an operator is not closable. Let X be the space of polynomial functions from [0,1] to ${\displaystyle \mathbb{R}}$ and Y the space of polynomial functions from [2,3] to ${\displaystyle \mathbb{R}}$. They are subspaces of C([0,1]) and C([2,3]) respectively, and so normed spaces. Define an operator T which takes the polynomial function x ↦ p(x) on [0,1] to the same function on [2,3]. As a consequence of the Stone–Weierstrass theorem, the graph of this operator is dense in ${\displaystyle X\times Y,}$ so this provides a sort of maximally discontinuous linear map (confer nowhere continuous function). Note that X is not complete here, as must be the case when there is such a constructible map.

Impact for dual spaces

The dual space of a topological vector space is the collection of continuous linear maps from the space into the underlying field. Thus the failure of some linear maps to be continuous for infinite-dimensional normed spaces implies that for these spaces, one needs to distinguish the algebraic dual space from the continuous dual space which is then a proper subset. It illustrates the fact that an extra dose of caution is needed in doing analysis on infinite-dimensional spaces as compared to finite-dimensional ones.

Beyond normed spaces

The argument for the existence of discontinuous linear maps on normed spaces can be generalized to all metrizable topological vector spaces, especially to all Fréchet spaces, but there exist infinite-dimensional locally convex topological vector spaces such that every functional is continuous.^[3] On the other hand, the Hahn–Banach theorem, which applies to all locally convex spaces, guarantees the existence of many continuous linear functionals, and so a large dual space. In fact, to every convex set, the Minkowski gauge associates a continuous linear functional. The upshot is that spaces with fewer convex sets have fewer functionals, and in the worst-case scenario, a space may have no functionals at all other than the zero functional. This is the case for the ${\displaystyle L^p(\mathbb{R},dx)}$ spaces with ${\displaystyle 0<p<1,}$ from which it follows that these spaces are nonconvex. Note that here is indicated the Lebesgue measure on the real line. There are other ${\displaystyle L^p}$ spaces with ${\displaystyle 0<p<1}$ which do have nontrivial dual spaces.

Another such example is the space of real-valued measurable functions on the unit interval with quasinorm given by

$${\displaystyle \Vert f\Vert ={\int }_I\frac{|f(x)|}{1+|f(x)|}dx.}$$

This non-locally convex space has a trivial dual space.

One can consider even more general spaces. For example, the existence of a homomorphism between complete separable metric groups can also be shown nonconstructively.

See also

• Finest locally convex topology – A vector space with a topology defined by convex open sets
• Sublinear function

References

1. 
   

Solovay, Robert M. (1970), "A model of set-theory in which every set of reals is Lebesgue measurable", Annals of Mathematics, Second Series, 92: 1–56, doi:10.2307/1970696, MR 0265151.

Schechter, Eric (1996), Handbook of Analysis and Its Foundations, Academic Press, p. 136, ISBN 9780080532998.

3. For example, the weak topology w.r.t. the space of all (algebraically) linear functionals.

• Constantin Costara, Dumitru Popa, Exercises in Functional Analysis, Springer, 2003. ISBN 1-4020-1560-7.
• Schechter, Eric, Handbook of Analysis and its Foundations, Academic Press, 1997. ISBN 0-12-622760-8.

v t e Functional analysis (topics – glossary)

v t e Topological vector spaces (TVSs)

Categories:

• Functional analysis
• Axiom of choice
• Functions and mappings

 https://en.wikipedia.org/wiki/Discontinuous_linear_map

https://en.wikipedia.org/wiki/Triangle_inequality

https://en.wikipedia.org/wiki/Uniform_norm