Absolute polar

Suppose that ${\displaystyle \langle X,Y\rangle }$ is a pairing. The polar or absolute polar of a subset ${\displaystyle A}$ of ${\displaystyle X}$ is the set:

$${\displaystyle {\begin{alignedat}{4}A^{\circ }:=&\left\{y\in Y~:~\sup _{a\in A}|\langle a,y\rangle |\leq 1\right\}~~~~&&\\[0.7ex]=&\left\{y\in Y~:~\sup |\langle A,y\rangle |\leq 1\right\}~~~~&&{\text{ where }}|\langle A,y\rangle |:=\{|\langle a,y\rangle |:a\in A\}\\[0.7ex]=&\left\{y\in Y~:~\langle A,y\rangle \subseteq B_{1}\right\}~~~~&&{\text{ where }}B_{1}:=\{s\in \mathbb {K}$$ :|s|\leq 1\}.\\[0.7ex]\end{alignedat}}}

where ${\displaystyle \langle A,y\rangle$ :=\{\langle a,y\rangle :a\in A\}} denotes the image of the set ${\displaystyle A}$ under the map ${\displaystyle \langle \cdot ,y\rangle :X\to \mathbb {K} }$ defined by ${\displaystyle x\mapsto \langle x,y\rangle .}$ If ${\displaystyle \operatorname {cobal} A}$ denotes the convex balanced hull of ${\displaystyle A,}$ which by definition is the smallest convex and balanced subset of ${\displaystyle X}$ that contains ${\displaystyle A,}$ then ${\displaystyle A^{\circ }=[\operatorname {cobal} A]^{\circ }.}$

This is an affine shift of the geometric definition; it has the useful characterization that the functional-analytic polar of the unit ball (in ${\displaystyle X}$) is precisely the unit ball (in ${\displaystyle Y}$).

The prepolar or absolute prepolar of a subset ${\displaystyle B}$ of ${\displaystyle Y}$ is the set:

$${\displaystyle {}^{\circ }B:=\left\{x\in X~:~\sup _{b\in B}|\langle x,b\rangle |\leq 1\right\}=\{x\in X~:~\sup |\langle x,B\rangle |\leq 1\}}$$

Very often, the prepolar of a subset ${\displaystyle B}$ of ${\displaystyle Y}$ is also called the polar or absolute polar of ${\displaystyle B}$ and denoted by ${\displaystyle B^{\circ }}$; in practice, this reuse of notation and of the word "polar" rarely causes any issues (such as ambiguity) and many authors do not even use the word "prepolar".

The bipolar of a subset ${\displaystyle A}$ of ${\displaystyle X,}$ often denoted by ${\displaystyle A^{\circ \circ },}$ is the set ${\displaystyle {}^{\circ }\left(A^{\circ }\right)}$; that is,

$${\displaystyle A^{\circ \circ }:={}^{\circ }\left(A^{\circ }\right)=\left\{x\in X~:~\sup _{y\in A^{\circ }}|\langle x,y\rangle |\leq 1\right\}.}$$

Real polar

The real polar of a subset ${\displaystyle A}$ of ${\displaystyle X}$ is the set:

$${\displaystyle A^{r}:=\left\{y\in Y~:~\sup _{a\in A}\operatorname {Re} \langle a,y\rangle \leq 1\right\}}$$

and the real prepolar of a subset ${\displaystyle B}$ of ${\displaystyle Y}$ is the set:

$${\displaystyle {}^{r}B:=\left\{x\in X~:~\sup _{b\in B}\operatorname {Re} \langle x,b\rangle \leq 1\right\}.}$$

As with the absolute prepolar, the real prepolar is usually called the real polar and is also denoted by ${\displaystyle B^{r}.}$^[2] It's important to note that some authors (e.g. [Schaefer 1999]) define "polar" to mean "real polar" (rather than "absolute polar", as is done in this article) and use the notation ${\displaystyle A^{\circ }}$ for it (rather than the notation ${\displaystyle A^{r}}$ that is used in this article and in [Narici 2011]).

The real bipolar of a subset ${\displaystyle A}$ of ${\displaystyle X,}$ sometimes denoted by ${\displaystyle A^{rr},}$ is the set ${\displaystyle {}^{r}\left(A^{r}\right)}$; it is equal to the ${\displaystyle \sigma (X,Y)}$-closure of the convex hull of ${\displaystyle A\cup \{0\}.}$^[2]

For a subset ${\displaystyle A}$ of ${\displaystyle X,}$ ${\displaystyle A^{r}}$ is convex, ${\displaystyle \sigma (Y,X)}$-closed, and contains ${\displaystyle A^{\circ }.}$^[2] In general, it is possible that ${\displaystyle A^{\circ }\neq A^{r}}$ but equality will hold if ${\displaystyle A}$ is balanced. Furthermore, ${\displaystyle A^{\circ }=\left(\operatorname {bal} \left(A^{r}\right)\right)}$ where ${\displaystyle \operatorname {bal} \left(A^{r}\right)}$ denotes the balanced hull of ${\displaystyle A^{r}.}$^[2]

Competing definitions

The definition of the "polar" of a set is not universally agreed upon. Although this article defined "polar" to mean "absolute polar", some authors define "polar" to mean "real polar" and other authors use still other definitions. No matter how an author defines "polar", the notation ${\displaystyle A^{\circ }}$ almost always represents their choice of the definition (so the meaning of the notation ${\displaystyle A^{\circ }}$ may vary from source to source). In particular, the polar of ${\displaystyle A}$ is sometimes defined as:

$${\displaystyle A^{|r|}:=\left\{y\in Y~:~\sup _{a\in A}|\operatorname {Re} \langle a,y\rangle |\leq 1\right\}}$$

where the notation ${\displaystyle A^{|r|}}$ is not standard notation.

We now briefly discuss how these various definitions relate to one another and when they are equivalent.

It is always the case that

$${\displaystyle A^{\circ }~\subseteq ~A^{|r|}~\subseteq ~A^{r}}$$

and if ${\displaystyle \langle \cdot ,\cdot \rangle }$ is real-valued (or equivalently, if ${\displaystyle X}$ and ${\displaystyle Y}$ are vector spaces over ${\displaystyle \mathbb {R} }$) then ${\displaystyle A^{\circ }=A^{|r|}.}$

If ${\displaystyle A}$ is a symmetric set (that is, ${\displaystyle -A=A}$ or equivalently, ${\displaystyle -A\subseteq A}$) then ${\displaystyle A^{|r|}=A^{r}}$ where if in addition ${\displaystyle \langle \cdot ,\cdot \rangle }$ is real-valued then ${\displaystyle A^{\circ }=A^{|r|}=A^{r}.}$

If ${\displaystyle X}$ and ${\displaystyle Y}$ are vector spaces over ${\displaystyle \mathbb {C} }$ (so that ${\displaystyle \langle \cdot ,\cdot \rangle }$ is complex-valued) and if ${\displaystyle iA\subseteq A}$ (where note that this implies ${\displaystyle -A=A}$ and ${\displaystyle iA=A}$), then

$${\displaystyle A^{\circ }\subseteq A^{|r|}=A^{r}\subseteq \left({\tfrac {1}{\sqrt {2}}}A\right)^{\circ }}$$

where if in addition ${\displaystyle e^{ir}A\subseteq A}$ for all real ${\displaystyle r}$ then ${\displaystyle A^{\circ }=A^{r}.}$

Thus for all of these definitions of the polar set of ${\displaystyle A}$ to agree, it suffices that ${\displaystyle sA\subseteq A}$ for all scalars ${\displaystyle s}$ of unit length^{[note 1]} (where this is equivalent to ${\displaystyle sA=A}$ for all unit length scalar ${\displaystyle s}$). In particular, all definitions of the polar of ${\displaystyle A}$ agree when ${\displaystyle A}$ is a balanced set (which is often, but not always, the case) so that often, which of these competing definitions is used is immaterial. However, these differences in the definitions of the "polar" of a set ${\displaystyle A}$ do sometimes introduce subtle or important technical differences when ${\displaystyle A}$ is not necessarily balanced.

Specialization for the canonical duality

Algebraic dual space

If ${\displaystyle X}$ is any vector space then let ${\displaystyle X^{\#}}$ denote the algebraic dual space of ${\displaystyle X,}$ which is the set of all linear functionals on ${\displaystyle X.}$ The vector space ${\displaystyle X^{\#}}$ is always a closed subset of the space ${\displaystyle \mathbb {K} ^{X}}$ of all ${\displaystyle \mathbb {K} }$-valued functions on ${\displaystyle X}$ under the topology of pointwise convergence so when ${\displaystyle X^{\#}}$ is endowed with the subspace topology, then ${\displaystyle X^{\#}}$ becomes a Hausdorff complete locally convex topological vector space (TVS). For any subset ${\displaystyle A\subseteq X,}$ let

$${\displaystyle {\begin{alignedat}{4}A^{\#}:=A^{\circ ,\#}:=&\left\{f\in X^{\#}~:~\sup _{a\in A}|f(a)|\leq 1\right\}&&\\[0.7ex]=&\left\{f\in X^{\#}~:~\sup |f(A)|\leq 1\right\}~~~~&&{\text{ where }}|f(A)|:=\{|f(a)|:a\in A\}\\[0.7ex]=&\left\{f\in X^{\#}~:~f(A)\subseteq B_{1}\right\}~~~&&{\text{ where }}B_{1}:=\{s\in \mathbb {K}$$ :|s|\leq 1\}.\\[0.7ex]\end{alignedat}}}

If ${\displaystyle A\subseteq B\subseteq X}$ are any subsets then ${\displaystyle B^{\#}\subseteq A^{\#}}$ and ${\displaystyle A^{\#}=[\operatorname {cobal} A]^{\#},}$ where ${\displaystyle \operatorname {cobal} A}$ denotes the convex balanced hull of ${\displaystyle A.}$ For any finite-dimensional vector subspace ${\displaystyle Y}$ of ${\displaystyle X,}$ let ${\displaystyle \tau _{Y}}$ denote the Euclidean topology on ${\displaystyle Y,}$ which is the unique topology that makes ${\displaystyle Y}$ into a Hausdorff topological vector space (TVS). If ${\displaystyle A_{\cup \operatorname {cl} \operatorname {Finite} }}$ denotes the union of all closures ${\displaystyle \operatorname {cl} _{\left(Y,\tau _{Y}\right)}(Y\cap A)}$ as ${\displaystyle Y}$ varies over all finite dimensional vector subspaces of ${\displaystyle X,}$ then ${\displaystyle A^{\#}=\left[A_{\cup \operatorname {cl} \operatorname {Finite} }\right]^{\#}}$ (see this footnote^{[note 2]} for an explanation). If ${\displaystyle A}$ is an absorbing subset of ${\displaystyle X}$ then by the Banach–Alaoglu theorem, ${\displaystyle A^{\#}}$ is a weak-* compact subset of ${\displaystyle X^{\#}.}$

If ${\displaystyle A\subseteq X}$ is any non-empty subset of a vector space ${\displaystyle X}$ and if ${\displaystyle Y}$ is any vector space of linear functionals on ${\displaystyle X}$ (that is, a vector subspace of the algebraic dual space of ${\displaystyle X}$) then the real-valued map

${\displaystyle |\,\cdot \,|_{A}\;:\,Y\,\to \,\mathbb {R} }$      defined by      ${\displaystyle \left|x^{\prime }\right|_{A}~:=~\sup \left|x^{\prime }(A)\right|~:=~\sup _{a\in A}\left|x^{\prime }(a)\right|}$

is a seminorm on ${\displaystyle Y.}$ If ${\displaystyle A=\varnothing }$ then by definition of the supremum, ${\displaystyle \,\sup \left|x^{\prime }(A)\right|=-\infty \,}$ so that the map ${\displaystyle \,|\,\cdot \,|_{\varnothing }=-\infty \,}$ defined above would not be real-valued and consequently, it would not be a seminorm.

Continuous dual space

Suppose that ${\displaystyle X}$ is a topological vector space (TVS) with continuous dual space ${\displaystyle X^{\prime }.}$ The important special case where ${\displaystyle Y:=X^{\prime }}$ and the brackets represent the canonical map:

$${\displaystyle \left\langle x,x^{\prime }\right\rangle$$ :=x^{\prime }(x)}

is now considered. The triple ${\displaystyle \left\langle X,X^{\prime }\right\rangle }$ is the called the canonical pairing associated with ${\displaystyle X.}$

The polar of a subset ${\displaystyle A\subseteq X}$ with respect to this canonical pairing is:

$${\displaystyle {\begin{alignedat}{4}A^{\circ }:=&\left\{x^{\prime }\in X^{\prime }~:~\sup _{a\in A}\left|x^{\prime }(a)\right|\leq 1\right\}~~~~&&{\text{ because }}\left\langle a,x^{\prime }\right\rangle$$ :=x^{\prime }(a)\\[0.7ex]=&\left\{x^{\prime }\in X^{\prime }~:~\sup \left|x^{\prime }(A)\right|\leq 1\right\}~~~~&&{\text{ where }}\left|x^{\prime }(A)\right|:=\left\{\left|x^{\prime }(a)\right|:a\in A\right\}\\[0.7ex]=&\left\{x^{\prime }\in X^{\prime }~:~x^{\prime }(A)\subseteq B_{1}\right\}~~~~&&{\text{ where }}B_{1}:=\{s\in \mathbb {K} :|s|\leq 1\}.\\[0.7ex]\end{alignedat}}}

For any subset ${\displaystyle A\subseteq X,}$ ${\displaystyle A^{\circ }=\left[\operatorname {cl} _{X}A\right]^{\circ }}$ where ${\displaystyle \operatorname {cl} _{X}A}$ denotes the closure of ${\displaystyle A}$ in ${\displaystyle X.}$

The Banach–Alaoglu theorem states that if ${\displaystyle A\subseteq X}$ is a neighborhood of the origin in ${\displaystyle X}$ then ${\displaystyle A^{\circ }=A^{\#}}$ and this polar set is a compact subset of the continuous dual space ${\displaystyle X^{\prime }}$ when ${\displaystyle X^{\prime }}$ is endowed with the weak-* topology (also known as the topology of pointwise convergence).

If ${\displaystyle A}$ satisfies ${\displaystyle sA\subseteq A}$ for all scalars ${\displaystyle s}$ of unit length then one may replace the absolute value signs by ${\displaystyle \operatorname {Re} }$ (the real part operator) so that:

$${\displaystyle {\begin{alignedat}{4}A^{\circ }=A^{r}:=&\left\{x^{\prime }\in X^{\prime }~:~\sup _{a\in A}\operatorname {Re} x^{\prime }(a)\leq 1\right\}\\[0.7ex]=&\left\{x^{\prime }\in X^{\prime }~:~\sup \operatorname {Re} x^{\prime }(A)\leq 1\right\}.\\[0.7ex]\end{alignedat}}}$$

The prepolar of a subset ${\displaystyle B}$ of ${\displaystyle Y=X^{\prime }}$ is:

$${\displaystyle {}^{\circ }B:=\left\{x\in X~:~\sup _{b^{\prime }\in B}\left|b^{\prime }(x)\right|\leq 1\right\}=\{x\in X:\sup |B(x)|\leq 1\}}$$

If ${\displaystyle B}$ satisfies ${\displaystyle sB\subseteq B}$ for all scalars ${\displaystyle s}$ of unit length then one may replace the absolute value signs with ${\displaystyle \operatorname {Re} }$ so that:

$${\displaystyle {}^{\circ }B=\left\{x\in X~:~\sup _{b^{\prime }\in B}\operatorname {Re} b^{\prime }(x)\leq 1\right\}=\{x\in X~:~\sup \operatorname {Re} B(x)\leq 1\}}$$

where ${\displaystyle B(x):=\left\{b^{\prime }(x)~:~b^{\prime }\in B\right\}.}$

The bipolar theorem characterizes the bipolar of a subset of a topological vector space.

If ${\displaystyle X}$ is a normed space and ${\displaystyle S}$ is the open or closed unit ball in ${\displaystyle X}$ (or even any subset of the closed unit ball that contains the open unit ball) then ${\displaystyle S^{\circ }}$ is the closed unit ball in the continuous dual space ${\displaystyle X^{\prime }}$ when ${\displaystyle X^{\prime }}$ is endowed with its canonical dual norm.