Dual system

 ( Functional Analysis ) 

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

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

---

Definition, notation, and conventions

---

Pairings A or pair over a field $\backslash mathbb\{K\}$ is a triple $(X,\; Y,\; b),$ which may also be denoted by $b(X,\; Y),$ consisting of two vector spaces $X$ and $Y$ over $\backslash mathbb\{K\}$ and a bilinear map $b\; :\; X\; \backslash times\; Y\; \backslash to\; \backslash mathbb\{K\}$ called the bilinear map associated with the pairing, or more simply called the pairing's map or its bilinear form. The examples here only describe when $\backslash mathbb\{K\}$ is either the $\backslash R$ or the $\backslash Complex$, but the mathematical theory is general.

For every $x\; \backslash in\; X$, define

                   & y && \mapsto &&\, b(x, y)
     

\end{alignat} and for every $y\; \backslash in\; Y,$ define

                   & x && \mapsto &&\, b(x, y).
     

\end{alignat} Every $b(x,\; \backslash ,\backslash cdot\backslash ,)$ is a linear functional on $Y$ and every $b(\backslash ,\backslash cdot\backslash ,,\; y)$ is a linear functional on $X$. Therefore both $$b(X,\; \backslash ,\backslash cdot\backslash ,)\; :=\; \backslash \{\; b(x,\; \backslash ,\backslash cdot\backslash ,)\; :\; x\; \backslash in\; X\; \backslash \}\; \backslash qquad\; \backslash text\{\; and\; \}\; \backslash qquad\; b(\backslash ,\backslash cdot\backslash ,,\; Y)\; :=\; \backslash \{\; b(\backslash ,\backslash cdot\backslash ,,\; y)\; :\; y\; \backslash in\; Y\; \backslash \},$$ form vector spaces of linear functionals.

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

---

Dual pairings A pairing $(X,\; Y,\; b)$ is called a , a , or a over $\backslash mathbb\{K\}$ if the bilinear form $b$ is non-degenerate, which means that it satisfies the following two separation axioms:

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

In this case $b$ is non-degenerate, and one can say that $b$ places $X$ and $Y$ in duality (or, redundantly but explicitly, in separated duality), and $b$ is called the duality pairing of the triple $(X,\; Y,\; b)$.

---

Total subsets A subset $S$ of $Y$ is called if for every $x\; \backslash in\; X$, $$b(x,\; s)\; =\; 0\; \backslash quad\; \backslash text\{\; for\; all\; \}\; s\; \backslash in\; S$$ implies $x\; =\; 0.$ A total subset of $X$ is defined analogously (see footnote).A subset $S$ of $X$ is total if for all $y\; \backslash in\; Y$, $$b(s,\; y)\; =\; 0\; \backslash quad\; \backslash text\{\; for\; all\; \}\; s\; \backslash in\; S$$ implies $y\; =\; 0$. Thus $X$ separates points of $Y$ if and only if $X$ is a total subset of $X$, and similarly for $Y$.

---

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

The orthogonal complement or annihilator of a subset $R\; \backslash subseteq\; X$ is $$R^\{\backslash perp\}\; :=\; \backslash \{\; y\; \backslash in\; Y\; :\; R\; \backslash perp\; y\; \backslash \}\; :=\; \backslash \{\; y\; \backslash in\; Y\; :\; b(R,\; y)\; =\; \backslash \{\; 0\; \backslash \}\; \backslash \}$$Thus $R$ is a total subset of $X$ if and only if $R^\backslash perp$ equals $\backslash \{0\backslash \}$.

---

Polar sets Given a triple $(X,\; Y,\; b)$ defining a pairing over $\backslash mathbb\{K\}$, the absolute polar set or polar set of a subset $A$ of $X$ is the set:$$A^\{\backslash circ\}\; :=\; \backslash left\backslash \{\; y\; \backslash in\; Y\; :\; \backslash sup\_\{x\; \backslash in\; A\}\; |b(x,\; y)|\; \backslash leq\; 1\; \backslash right\backslash \}.$$Symmetry, the absolute polar set or polar set of a subset $B$ of $Y$ is denoted by $B^\{\backslash circ\}$ and defined by $$B^\{\backslash circ\}\; :=\; \backslash left\backslash \{\; x\; \backslash in\; X\; :\; \backslash sup\_\{y\; \backslash in\; B\}\; |b(x,\; y)|\; \backslash leq\; 1\; \backslash right\backslash \}.$$

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

The polar $B^\{\backslash circ\}$ is necessarily a Convex set set containing $0\; \backslash in\; Y$ where if $B$ is balanced then so is $B^\{\backslash circ\}$ and if $B$ is a vector subspace of $X$ then so too is $B^\{\backslash circ\}$ a vector subspace of $Y.$

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

---

Dual definitions and results Given a pairing $(X,\; Y,\; b),$ define a new pairing $(Y,\; X,\; d)$ where $d(y,\; x)\; :=\; b(x,\; y)$ for all $x\; \backslash in\; X$ and $y\; \backslash in\; Y$.

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

: Given any definition for a pairing $(X,\; Y,\; b),$ one obtains a by applying it to the pairing $(Y,\; X,\; d).$ These conventions also apply to theorems.

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

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

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

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

---

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

---

Examples

---

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

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

---

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

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

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

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

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

---

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

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

---

Polars and duals of TVSs The following result shows that the continuous linear functionals on a TVS are exactly those linear functionals that are bounded on a neighborhood of the origin.

---

Inner product spaces and complex conjugate spaces A pre-Hilbert space $(H,\; \backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle)$ is a dual pairing if and only if $H$ is vector space over $\backslash R$ or $H$ has dimension $0.$ Here it is assumed that the sesquilinear form $\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle$ is conjugate homogeneous in its second coordinate and homogeneous in its first coordinate.

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

Suppose that $(H,\; \backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle)$ is a complex pre-Hilbert space with scalar multiplication denoted as usual by juxtaposition or by a dot $\backslash cdot.$ Define the map $$\backslash ,\backslash cdot\backslash ,\; \backslash perp\; \backslash ,\backslash cdot\backslash ,\; :\; \backslash Complex\; \backslash times\; H\; \backslash to\; H\; \backslash quad\; \backslash text\{\; by\; \}\; \backslash quad\; c\; \backslash perp\; x\; :=\; \backslash overline\{c\}\; x,$$ where the right-hand side uses the scalar multiplication of $H.$ Let $\backslash overline\{H\}$ denote the complex conjugate vector space of $H,$ where $\backslash overline\{H\}$ denotes the additive group of $(H,\; +)$ (so vector addition in $\backslash overline\{H\}$ is identical to vector addition in $H$) but with scalar multiplication in $\backslash overline\{H\}$ being the map $\backslash ,\backslash cdot\backslash ,\; \backslash perp\; \backslash ,\backslash cdot\backslash ,$ (instead of the scalar multiplication that $H$ is endowed with).

The map $b\; :\; H\; \backslash times\; \backslash overline\{H\}\; \backslash to\; \backslash Complex$ defined by $b(x,\; y)\; :=\; \backslash langle\; x,\; y\; \backslash rangle$ is linear in both coordinatesThat $b$ is linear in its first coordinate is obvious. Suppose $c$ is a scalar. Then $b(x,\; c\; \backslash perp\; y)\; =\; b\backslash left(x,\; \backslash overline\{c\}\; y\backslash right)\; =\; \backslash langle\; x,\; \backslash overline\{c\}\; y\; \backslash rangle\; =\; c\; \backslash langle\; x,\; y\; \backslash rangle\; =\; c\; b(x,\; y),$ which shows that $b$ is linear in its second coordinate. and so $\backslash left(H,\; \backslash overline\{H\},\; \backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle\backslash right)$ forms a dual pairing.

---

Other examples

• Suppose $X\; =\; \backslash R^2,$ $Y\; =\; \backslash R^3,$ and for all $\backslash left(x\_1,\; y\_1\backslash right)\; \backslash in\; X\; \backslash text\{\; and\; \}\; \backslash left(x\_2,\; y\_2,\; z\_2\backslash right)\; \backslash in\; Y,$ let $$b\backslash left(\backslash left(x\_1,\; y\_1\backslash right),\; \backslash left(x\_2,\; y\_2,\; z\_2\backslash right)\backslash right)\; :=\; x\_1\; x\_2\; +\; y\_1\; y\_2.$$ Then $(X,\; Y,\; b)$ is a pairing such that $X$ distinguishes points of $Y,$ but $Y$ does not distinguish points of $X.$ Furthermore, $X^\{\backslash perp\}\; :=\; \backslash \{y\; \backslash in\; Y\; :\; X\; \backslash perp\; y\backslash \}\; =\; \backslash \{(0,\; 0,\; z)\; :\; z\; \backslash in\; \backslash R\backslash \}.$
• Let $X\; :=\; L^p(\backslash mu),$ $Y\; :=\; L^q(\backslash mu)$ (where $q$ is such that $\backslash tfrac\{1\}\{p\}\; +\; \backslash tfrac\{1\}\{q\}\; =\; 1$), and $b(f,\; g)\; :=\; \backslash int\; f\; g\; \backslash ,\; \backslash mathrm\{d\}\backslash mu.$ Then $(X,\; Y,\; b)$ is a dual system.
• Let $X$ and $Y$ be vector spaces over the same field $\backslash mathbb\{K\}.$ Then the bilinear form $b\backslash left(x\; \backslash otimes\; y,\; x^*\; \backslash otimes\; y^*\backslash right)\; =\; \backslash left\backslash langle\; x^\{\backslash prime\},\; x\; \backslash right\backslash rangle\; \backslash left\backslash langle\; y^\{\backslash prime\},\; y\; \backslash right\backslash rangle$ places $X\; \backslash times\; Y$ and $X^\{\backslash \#\}\; \backslash times\; Y^\{\backslash \#\}$ in duality.
• A sequence space $X$ and its beta dual $Y\; :=\; X^\{\backslash beta\}$ with the bilinear map defined as $\backslash langle\; x,\; y\; \backslash rangle\; :=\; \backslash sum\_\{i=1\}^\{\backslash infty\}\; x\_i\; y\_i$ for $x\; \backslash in\; X,$ $y\; \backslash in\; X^\{\backslash beta\}$ forms a dual system.

---

Weak topology Suppose that $(X,\; Y,\; b)$ is a pairing of Vector space over $\backslash mathbb\{K\}.$ If $S\; \backslash subseteq\; Y$ then the weak topology on $X$ induced by $S$ (and $b$) is the weakest TVS topology on $X,$ denoted by $\backslash sigma(X,\; S,\; b)$ or simply $\backslash sigma(X,\; S),$ making each map $b(\backslash ,\backslash cdot\backslash ,,\; y)\; :\; X\; \backslash to\; \backslash mathbb\{K\}$ continuous as a function of $x$ for every $y\; \backslash in\; S$. If $S$ is not clear from context then it should be assumed to be all of $Y,$ in which case it is called the weak topology on $X$ (induced by $Y$). The notation $X\_\{\backslash sigma(X,\; S,\; b)\},$ $X\_\{\backslash sigma(X,\; S)\},$ or (if no confusion could arise) simply $X\_\{\backslash sigma\}$ is used to denote $X$ endowed with the weak topology $\backslash sigma(X,\; S,\; b).$ Importantly, the weak topology depends on the function $b,$ the usual topology on $\backslash Complex,$ and $X$'s vector space structure but on the algebraic structures of $Y.$

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

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

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

If $(X,\; Y,\; b)$ is a pairing and $N$ is a proper vector subspace of $Y$ such that $(X,\; N,\; b)$ is a dual pair, then $\backslash sigma(X,\; N,\; b)$ is strictly coarser than $\backslash sigma(X,\; Y,\; b).$

---

Bounded subsets A subset $S$ of $X$ is $\backslash sigma(X,\; Y,\; b)$-bounded if and only if where $|b(S,\; y)|\; :=\; \backslash \{\; b(s,\; y)\; :\; s\; \backslash in\; S\; \backslash \}.$

---

Hausdorffness If $(X,\; Y,\; b)$ is a pairing then the following are equivalent:

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

---

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

Consequently, the continuous dual space of $(X,\; \backslash sigma(X,\; Y,\; b))$ is $$(X,\; \backslash sigma(X,\; Y,\; b))^\{\backslash prime\}\; =\; b(\backslash ,\backslash cdot\backslash ,,\; Y)\; :=\; \backslash left\backslash \{\; b(\backslash ,\backslash cdot\backslash ,,\; y)\; :\; y\; \backslash in\; Y\; \backslash right\backslash \}.$$

With respect to the canonical pairing, if $X$ is a TVS whose continuous dual space $X^\{\backslash prime\}$ separates points on $X$ (i.e. such that $\backslash left(X,\; \backslash sigma\backslash left(X,\; X^\{\backslash prime\}\backslash right)\backslash right)$ is Hausdorff, which implies that $X$ is also necessarily Hausdorff) then the continuous dual space of $\backslash left(X^\{\backslash prime\},\; \backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)\backslash right)$ is equal to the set of all "evaluation at a point $x$" maps as $x$ ranges over $X$ (i.e. the map that send $x^\{\backslash prime\}\; \backslash in\; X^\{\backslash prime\}$ to $x^\{\backslash prime\}(x)$). This is commonly written as $$\backslash left(X^\{\backslash prime\},\; \backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)\backslash right)^\{\backslash prime\}\; =\; X\; \backslash qquad\; \backslash text\{\; or\; \}\; \backslash qquad\; \backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\}\; =\; X.$$ This very important fact is why results for polar topologies on continuous dual spaces, such as the strong dual topology $\backslash beta\backslash left(X^\{\backslash prime\},\; X\backslash right)$ on $X^\{\backslash prime\}$ for example, can also often be applied to the original TVS $X$; for instance, $X$ being identified with $\backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\}$ means that the topology $\backslash beta\backslash left(\backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\},\; X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)$ on $\backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\}$ can instead be thought of as a topology on $X.$ Moreover, if $X^\{\backslash prime\}$ is endowed with a topology that is finer than $\backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)$ then the continuous dual space of $X^\{\backslash prime\}$ will necessarily contain $\backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\}$ as a subset. So for instance, when $X^\{\backslash prime\}$ is endowed with the strong dual topology (and so is denoted by $X^\{\backslash prime\}\_\{\backslash beta\}$) then $$\backslash left(X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)^\{\backslash prime\}\; ~\backslash supseteq~\; \backslash left(X^\{\backslash prime\}\_\{\backslash sigma\}\backslash right)^\{\backslash prime\}\; ~=~\; X$$ which (among other things) allows for $X$ to be endowed with the subspace topology induced on it by, say, the strong dual topology $\backslash beta\backslash left(\backslash left(X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)^\{\backslash prime\},\; X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)$ (this topology is also called the strong bidual topology and it appears in the theory of : the Hausdorff locally convex TVS $X$ is said to be if $\backslash left(X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)^\{\backslash prime\}\; =\; X$ and it will be called if in addition the strong bidual topology $\backslash beta\backslash left(\backslash left(X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)^\{\backslash prime\},\; X^\{\backslash prime\}\_\{\backslash beta\}\backslash right)$ on $X$ is equal to $X$'s original/starting topology).

---

Orthogonals, quotients, and subspaces If $(X,\; Y,\; b)$ is a pairing then for any subset $S$ of $X$:

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

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

---

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

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

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

---

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

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

---

Polars and the weak topology If $X$ is a locally convex space and if $H$ is a subset of the continuous dual space $X^\{\backslash prime\},$ then $H$ is $\backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)$-bounded if and only if $H\; \backslash subseteq\; B^\{\backslash circ\}$ for some Barrelled space $B$ in $X.$

The following results are important for defining polar topologies.

If $(X,\; Y,\; b)$ is a pairing and $A\; \backslash subseteq\; X,$ then:

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

If $(X,\; Y,\; b)$ is a pairing and $\backslash tau$ is a locally convex topology on $X$ that is consistent with duality, then a subset $B$ of $X$ is a Barreled space in $(X,\; \backslash tau)$ if and only if $B$ is the Polar set of some $\backslash sigma(Y,\; X,\; b)$-bounded subset of $Y.$

---

Transposes

---

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

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

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

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

called the transpose or adjoint of $F$ with respect to $(X,\; Y,\; b)$ and $(W,\; Z,\; c)$ (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 $\{\}^t\; F$ to be well-defined. For every $z\; \backslash in\; Z,$ the defining condition for $\{\}^t\; F(z)$ is $$c(F(\backslash ,\backslash cdot\backslash ,),\; z)\; =\; b\backslash left(\backslash ,\backslash cdot\backslash ,,\; \{\}^t\; F(z)\backslash right),$$ that is, $$c(F(x),\; z)\; =\; b\backslash left(x,\; \{\}^t\; F(z)\backslash right)$$ for all $x\; \backslash in\; X.$

By the conventions mentioned at the beginning of this article, this also defines the transpose of linear maps of the form $Z\; \backslash to\; Y,$If $G\; :\; Z\; \backslash to\; Y$ is a linear map then $G$'s transpose, $\{\}^t\; G\; :\; X\; \backslash to\; W,$ is well-defined if and only if $Z$ distinguishes points of $W$ and $b(X,\; G(\backslash ,\backslash cdot\backslash ,))\; \backslash subseteq\; c(W,\; \backslash ,\backslash cdot\backslash ,).$ In this case, for each $x\; \backslash in\; X,$ the defining condition for $\{\}^t\; G(x)$ is: $c(x,\; G(\backslash ,\backslash cdot\backslash ,))\; =\; c\backslash left(\{\}^t\; G(x),\; \backslash ,\backslash cdot\backslash ,\backslash right).$ $X\; \backslash to\; Z,$If $H\; :\; X\; \backslash to\; Z$ is a linear map then $H$'s transpose, $\{\}^t\; H\; :\; W\; \backslash to\; Y,$ is well-defined if and only if $X$ distinguishes points of $Y$ and $c(W,\; H(\backslash ,\backslash cdot\backslash ,))\; \backslash subseteq\; b(\backslash ,\backslash cdot\backslash ,,\; Y).$ In this case, for each $w\; \backslash in\; W,$ the defining condition for $\{\}^t\; H(w)$ is: $c(w,\; H(\backslash ,\backslash cdot\backslash ,))\; =\; b\backslash left(\backslash ,\backslash cdot\backslash ,,\; \{\}^t\; H(w)\backslash right).$ $W\; \backslash to\; Y,$If $H\; :\; W\; \backslash to\; Y$ is a linear map then $H$'s transpose, $\{\}^t\; H\; :\; X\; \backslash to\; Q,$ is well-defined if and only if $W$ distinguishes points of $Z$ and $b(X,\; H(\backslash ,\backslash cdot\backslash ,))\; \backslash subseteq\; c(\backslash ,\backslash cdot\backslash ,,\; Z).$ In this case, for each $x\; \backslash in\; X,$ the defining condition for $\{\}^t\; H(x)$ is: $c(x,\; H(\backslash ,\backslash cdot\backslash ,))\; =\; b\backslash left(\backslash ,\backslash cdot\backslash ,,\; \{\}^t\; H(x)\backslash right).$ $Y\; \backslash to\; W,$If $H\; :\; Y\; \backslash to\; W$ is a linear map then $H$'s transpose, $\{\}^t\; H\; :\; Z\; \backslash to\; X,$ is well-defined if and only if $Y$ distinguishes points of $X$ and $c(H(\backslash ,\backslash cdot\backslash ,),\; Z)\; \backslash subseteq\; b(X,\; \backslash ,\backslash cdot\backslash ,).$ In this case, for each $z\; \backslash in\; Z,$ the defining condition for $\{\}^t\; H(z)$ is: $c(H(\backslash ,\backslash cdot\backslash ,),\; z)\; =\; b\backslash left(\{\}^t\; H(z),\; \backslash ,\backslash cdot\backslash ,\backslash right)$ etc. (see footnote).

---

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

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

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

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

---

Weak continuity A linear map $F\; :\; X\; \backslash to\; W$ is weakly continuous (with respect to $(X,\; Y,\; b)$ and $(W,\; Z,\; c)$) if $F\; :\; (X,\; \backslash sigma(X,\; Y,\; b))\; \backslash to\; (W,\; (W,\; Z,\; c))$ is continuous.

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

---

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

---

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

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

In particular, with respect to the canonical duality, if $Y$ is a vector subspace of $X^\{\backslash \#\}$ such that $Y$ separates points of $X,$ then $(Y,\; \backslash sigma(Y,\; X))$ is complete if and only if $Y\; =\; X^\{\backslash \#\}.$ Said differently, there does exist a proper vector subspace $Y\; \backslash neq\; X^\{\backslash \#\}$ of $X^\{\backslash \#\}$ such that $(X,\; \backslash sigma(X,\; Y))$ is Hausdorff and $Y$ is complete in the weak-* topology (i.e. the topology of pointwise convergence). Consequently, when the continuous dual space $X^\{\backslash prime\}$ of a Hausdorff space locally convex TVS $X$ is endowed with the weak-* topology, then $X^\{\backslash prime\}\_\{\backslash sigma\}$ is complete if and only if $X^\{\backslash prime\}\; =\; X^\{\backslash \#\}$ (that is, if and only if linear functional on $X$ is continuous).

---

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

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

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

---

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

If $X\; =\; Y\; =\; \backslash mathbb\{K\}^n$ for some integer $n,$ $\backslash mathcal\{E\}\; =\; \backslash left\backslash \{\; e\_1,\; \backslash ldots,\; e\_n\backslash right\backslash \}$ is a basis for $X$ with dual basis $\backslash mathcal\{E\}^\{\backslash prime\}\; =\; \backslash left\backslash \{\; e\_1^\{\backslash prime\},\; \backslash ldots,\; e\_n^\{\backslash prime\}\; \backslash right\backslash \},$ $F\; :\; \backslash mathbb\{K\}^n\; \backslash to\; \backslash mathbb\{K\}^n$ is a linear operator, and the matrix representation of $F$ with respect to $\backslash mathcal\{E\}$ is $M\; :=\; \backslash left(f\_\{i,j\}\backslash right),$ then the transpose of $M$ is the matrix representation with respect to $\backslash mathcal\{E\}^\{\backslash prime\}$ of $F^\{\backslash \#\}.$

---

Weak continuity and openness Suppose that $\backslash left\backslash langle\; X,\; Y\; \backslash right\backslash rangle$ and $\backslash langle\; W,\; Z\; \backslash rangle$ are canonical pairings (so $Y\; \backslash subseteq\; X^\{\backslash \#\}$ and $Z\; \backslash subseteq\; W^\{\backslash \#\}$) that are dual systems and let $F\; :\; X\; \backslash to\; W$ be a linear map. Then $F\; :\; X\; \backslash to\; W$ is weakly continuous if and only if it satisfies any of the following equivalent conditions:

1. $F\; :\; (X,\; \backslash sigma(X,\; Y))\; \backslash to\; (W,\; \backslash sigma(W,\; Z))$ is continuous.
2. $F^\{\backslash \#\}(Z)\; \backslash subseteq\; Y$
3. the transpose of F, $\{\}^t\; F\; :\; Z\; \backslash to\; Y,$ with respect to $\backslash left\backslash langle\; X,\; Y\; \backslash right\backslash rangle$ and $\backslash langle\; W,\; Z\; \backslash rangle$ is well-defined.

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

A map $g\; :\; A\; \backslash to\; B$ between topological spaces is relatively open if $g\; :\; A\; \backslash to\; \backslash operatorname\{Im\}\; g$ is an , where $\backslash operatorname\{Im\}\; g$ is the range of $g.$

Suppose that $\backslash langle\; X,\; Y\; \backslash rangle$ and $\backslash langle\; W,\; Z\; \backslash rangle$ are dual systems and $F\; :\; X\; \backslash to\; W$ is a weakly continuous linear map. Then the following are equivalent:

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

Furthermore,

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

---

Transpose of a map between TVSs The transpose of map between two TVSs is defined if and only if $F$ is weakly continuous.

If $F\; :\; X\; \backslash to\; Y$ is a linear map between two Hausdorff locally convex topological vector spaces, then:

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

---

Metrizability and separability Let $X$ be a locally convex space with continuous dual space $X^\{\backslash prime\}$ and let $K\; \backslash subseteq\; X^\{\backslash prime\}.$

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

---

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

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

---

Polar topologies Given a collection $\backslash mathcal\{G\}$ of subsets of $X$, the polar topology on $Y$ determined by $\backslash mathcal\{G\}$ (and $b$) or the $\backslash mathcal\{G\}$-topology on $Y$ is the unique topological vector space (TVS) topology on $Y$ for which $$\backslash left\backslash \{\; r\; G^\{\backslash circ\}\; :\; G\; \backslash in\; \backslash mathcal\{G\},\; r\; >\; 0\; \backslash right\backslash \}$$ forms a subbasis of neighborhoods at the origin. When $Y$ is endowed with this $\backslash mathcal\{G\}$-topology then it is denoted by Y_{$\backslash mathcal\{G\}$}. Every polar topology is necessarily locally convex. When $\backslash mathcal\{G\}$ is a directed set with respect to subset inclusion (i.e. if for all $G,\; K\; \backslash in\; \backslash mathcal\{G\}$ there exists some $K\; \backslash in\; \backslash mathcal\{G\}$ such that $G\; \backslash cup\; H\; \backslash subseteq\; K$) then this neighborhood subbasis at 0 actually forms a neighborhood basis at 0.

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

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

finite subsets of $X$ (or $\backslash sigma(X,\; Y,\; b)$-closed disked hulls of finite subsets of $X$)	$\backslash sigma(X,\; Y,\; b)$ $s(X,\; Y,\; b)$	pointwise/simple convergence	Weak topology
$\backslash sigma(X,\; Y,\; b)$-compact disks	$\backslash tau(X,\; Y,\; b)$		Mackey topology
$\backslash sigma(X,\; Y,\; b)$-compact convex subsets	$\backslash gamma(X,\; Y,\; b)$	compact convex convergence
$\backslash sigma(X,\; Y,\; b)$-compact subsets (or balanced $\backslash sigma(X,\; Y,\; b)$-compact subsets)	$c(X,\; Y,\; b)$	compact convergence
$\backslash sigma(X,\; Y,\; b)$-bounded subsets	$b(X,\; Y,\; b)$ $\backslash beta(X,\; Y,\; b)$	bounded convergence	strong topology Strongest polar topology

---

Definitions involving polar topologies Continuity

A linear map $F\; :\; X\; \backslash to\; W$ is Mackey continuous (with respect to $(X,\; Y,\; b)$ and $(W,\; Z,\; c)$) if $F\; :\; (X,\; \backslash tau(X,\; Y,\; b))\; \backslash to\; (W,\; \backslash tau(W,\; Z,\; c))$ is continuous.

A linear map $F\; :\; X\; \backslash to\; W$ is strongly continuous (with respect to $(X,\; Y,\; b)$ and $(W,\; Z,\; c)$) if $F\; :\; (X,\; \backslash beta(X,\; Y,\; b))\; \backslash to\; (W,\; \backslash beta(W,\; Z,\; c))$ is continuous.

---

Bounded subsets A subset of $X$ is weakly bounded (resp. Mackey bounded, strongly bounded) if it is bounded in $(X,\; \backslash sigma(X,\; Y,\; b))$ (resp. bounded in $(X,\; \backslash tau(X,\; Y,\; b)),$ bounded in $(X,\; \backslash beta(X,\; Y,\; b))$).

---

Topologies compatible with a pair If $(X,\; Y,\; b)$ is a pairing over $\backslash mathbb\{K\}$ and $\backslash mathcal\{T\}$ is a vector topology on $X$ then $\backslash mathcal\{T\}$ is a topology of the pairing and that it is compatible (or consistent) with the pairing $(X,\; Y,\; b)$ if it is locally convex and if the continuous dual space of $\backslash left(X,\; \backslash mathcal\{T\}\backslash right)\; =\; b(\backslash ,\backslash cdot\backslash ,,\; Y).$Of course, there is an analogous definition for topologies on $Y$ to be "compatible it a pairing" but this article will only deal with topologies on $X.$ If $X$ distinguishes points of $Y$ then by identifying $Y$ as a vector subspace of $X$'s algebraic dual, the defining condition becomes: $\backslash left(X,\; \backslash mathcal\{T\}\backslash right)^\{\backslash prime\}\; =\; Y.$ Some authors (e.g. Trèves and Schaefer) require that a topology of a pair also be Hausdorff, which it would have to be if $Y$ distinguishes the points of $X$ (which these authors assume).

The weak topology $\backslash sigma(X,\; Y,\; b)$ is compatible with the pairing $(X,\; Y,\; b)$ (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 $N$ is a normed space that is not Reflexive space then the usual norm topology on its continuous dual space is compatible with the duality $\backslash left(N^\{\backslash prime\},\; N\backslash right).$

---

Mackey–Arens theorem The following is one of the most important theorems in duality theory.

It follows that the Mackey topology $\backslash tau(X,\; Y,\; b),$ which recall is the polar topology generated by all $\backslash sigma(X,\; Y,\; b)$-compact disks in $Y,$ is the strongest locally convex topology on $X$ that is compatible with the pairing $(X,\; Y,\; b).$ 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's theorem, barrels, and closed convex sets If $X$ is a TVS (over $\backslash Reals$ or $\backslash Complex$) then a half-space is a set of the form $\backslash \{\; x\; \backslash in\; X\; :\; f(x)\; \backslash leq\; r\; \backslash \}$ for some real $r$ and some continuous linear functional $f$ on $X.$

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

The following theorem shows that Barrelled space (i.e. closed Absorbing set disks) are exactly the polars of weakly bounded subsets.

If $X$ is a topological vector space, then:

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

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.

---

Space of finite sequences Let $X$ denote the space of all sequences of scalars $r\_\{\backslash bull\}\; =\; \backslash left(r\_i\backslash right)\_\{i=1\}^\{\backslash infty\}$ such that $r\_i\; =\; 0$ for all sufficiently large $i.$ Let $Y\; =\; X$ and define a bilinear map $b\; :\; X\; \backslash times\; X\; \backslash to\; \backslash mathbb\{K\}$ by $$b\backslash left(r\_\{\backslash bull\},\; s\_\{\backslash bull\}\backslash right)\; :=\; \backslash sum\_\{i=1\}^\{\backslash infty\}\; r\_i\; s\_i.$$ Then $\backslash sigma(X,\; X,\; b)\; =\; \backslash tau(X,\; X,\; b).$ Moreover, a subset $T\; \backslash subseteq\; X$ is $\backslash sigma(X,\; X,\; b)$-bounded (resp. $\backslash beta(X,\; X,\; b)$-bounded) if and only if there exists a sequence $m\_\{\backslash bull\}\; =\; \backslash left(m\_i\backslash right)\_\{i=1\}^\{\backslash infty\}$ of positive real numbers such that $\backslash left|t\_i\backslash right|\; \backslash leq\; m\_i$ for all $t\_\{\backslash bull\}\; =\; \backslash left(t\_i\backslash right)\_\{i=1\}^\{\backslash infty\}\; \backslash in\; T$ and all indices $i$ (resp. and $m\_\{\backslash bull\}\; \backslash in\; X$).

It follows that there are weakly bounded (that is, $\backslash sigma(X,\; X,\; b)$-bounded) subsets of $X$ that are not strongly bounded (that is, not $\backslash beta(X,\; X,\; b)$-bounded).

---

See also

---

Notes

---

Bibliography

• Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, Vol. 1, Functional Analysis, Section III.3. Academic Press, San Diego, 1980. .

---

External links

• Duality Theory

Post Comment