16.2 Polars
Definition 16.2.1 (Real Polar).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
is the real polar of $A$.
Definition 16.2.2 (Absolute Polar).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
is the absolute polar of $A$.
Proposition 16.2.3.label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
- (1)
$\emptyset^{\circ} = F$ and $E^{\circ} = \bracs{0}$.
- (2)
For any $\alpha \in K \setminus \bracs{0}$ and $A \subset E$, $(\alpha A)^{\circ} = \alpha^{-1}\cdot A^{\circ}$.
- (3)
For any $\seqi{A}\subset E$, $\paren{\bigcup_{i \in I}A_i}^{\circ} = \bigcap_{i \in I}A_{i}^{\circ}$.
- (4)
For any $A \subset B \subset E$, $A^{\circ} \supset B^{\circ}$.
- (5)
For any saturated ideal $\sigma \subset \mathfrak{B}(E, \sigma(E, F))$, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology on $F$.
Proof. (2): For any $\lambda \in K \setminus \bracs{0}$ and $A \subset E$,
(5): Let $S \in \sigma$, then
Since $\sigma$ is saturated, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology.$\square$
Proposition 16.2.4 ([IV.1.4, SW99]).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
- (1)
$A^{\circ}$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$.
- (2)
If $A$ is circled, then so is $A^{\circ}$.
- (3)
If $A$ is a subspace of $E$, then
\[A^{\circ} = A^{\perp} = \bracs{y \in F| \dpn{x, y}{E} = 0 \forall x \in A}\]and $A^{\circ}$ is a subspace of $F$.
Proof. (1): For each $x \in E$,
is the sublevel set of a continuous $\real$-linear functional, so it is $\sigma(F, E)$-closed and convex. Since $A^{\circ} = \bigcap_{x \in A}\bracs{x}^{\circ}$, $A$ is also $\sigma(F, E)$-closed and convex.
(2): If $A$ is circled, then by Proposition 16.2.3,
For any $y \in A^{\circ}$ and $\alpha \in K \setminus \bracs{0}$ with $|\alpha| \le 1$. Since $y \in \alpha^{-1}A^{\circ}$, $\alpha y \in A^{\circ}$, so $A^{\circ}$ is circled.$\square$
Proposition 16.2.5.label Let $E$ be a TVS over $K \in \RC$, $\dpn{E, E^*}{E}$ be the canonical duality, and $A \subset E^{*}$, then the following are equivalent
- (1)
$A$ is equicontinuous.
- (2)
$A^{\circ} \in \cn_{E}(0)$.
Proof. By Proposition 10.13.1, $A$ is equicontinuous if and only if
(1) $\Rightarrow$ (2): $A^{\circ} \supset \bigcap_{\phi \in A}\phi^{-1}(B_{K}(0, 1))$.
(2) $\Rightarrow$ (1): Since $A^{\circ} \in \cn_{E}(0)$, there exists $V \in \cn_{E}(0)$ circled with $V \subset A^{\circ}$, so $V \subset \bigcap_{\phi \in A}\phi^{-1}(B_{K}(0, 1))$, and $\bigcap_{\phi \in A}\phi^{-1}(B_{K}(0, 1)) \in \cn_{E}(0)$.$\square$
Theorem 16.2.6 (Bipolar Theorem).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. For each $A \subset F$,
with respect to the $\sigma(E, F)$-topology.
Proof, [IV.1.5, SW99]. By Proposition 16.2.4, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ}\supset A$, it is sufficient to show that $A^{\circ\circ}\subset \ol{\conv}(A \cup \bracs{0})$.
Let $x_{0} \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the [Hahn-Banach Theorem]theorem:hahn-banach-geometric-2, there exists $\phi: E \to \real$ such that:
- (1)
$\phi$ is $\sigma(E, F)$-continuous.
- (2)
$\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_{0}) > 1$.
If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by Proposition 10.4.1, the mapping $\Phi(x) = \phi(x) - i\phi(ix)$ is a $\sigma(E; F)$-continuous $K$-linear map on $E$ such that $\text{Re}(\Phi) = \phi$. By Lemma 16.1.3, there exists $y \in F$ such that $\dpn{x, y}{\lambda}= \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,
Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda}> 1$, so $x_{0} \not\in A^{\circ\circ}$.$\square$