\section{Polars}
\label{section:polar}

\begin{definition}[Real Polar]
\label{definition:real-polar}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
    \[
    A^\circ = \bracsn{y \in F| \text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in A}
    \]

    is the \textbf{real polar} of $A$. 
\end{definition}

\begin{definition}[Absolute Polar]
\label{definition:absolute-polar}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
    \[
    A^\square = \bracsn{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A}
    \]

    is the \textbf{absolute polar} of $A$. 
\end{definition}

\begin{proposition}
\label{proposition:polar-gymnastics}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
    \begin{enumerate}
        \item $\emptyset^\circ = F$ and $E^\circ = \bracs{0}$. 
        \item For any $\alpha \in K \setminus \bracs{0}$ and $A \subset E$, $(\alpha A)^\circ = \alpha^{-1} \cdot A^\circ$.
        \item For any $\seqi{A} \subset E$, $\paren{\bigcup_{i \in I}A_i}^\circ = \bigcap_{i \in I}A_i^\circ$. 
        \item For any $A \subset B \subset E$, $A^\circ \supset B^\circ$. 
        \item 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 $F$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (2): For any $\lambda \in K \setminus \bracs{0}$ and $A \subset E$, 
    \begin{align*}
        (\lambda A)^\circ &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
        &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
        &= \bracs{y \in F|\text{Re}\dpn{x, \alpha y}{\lambda} \le 1 \forall x \in A} \\
        &= \bracs{y \in F|\text{Re}\dpn{\alpha x,  y}{\lambda} \le 1 \forall x \in A} \\
        &= \bracs{\alpha^{-1} y \in F|\text{Re}\dpn{x,  y}{\lambda} \le 1 \forall x \in A} \\
        &= \alpha^{-1} \cdot A^\circ
    \end{align*}

    (5): Let $S \in \sigma$, then
    \[
    (\aconv(S))^\circ = \bracs{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A} \subset S^\circ
    \]

    Since $\sigma$ is saturated, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology.     
\end{proof}

\begin{proposition}[{{\cite[IV.1.4]{SchaeferWolff}}}]
\label{proposition:polar-properties}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
    \begin{enumerate}
        \item $A^\circ$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$. 
        \item If $A$ is circled, then so is $A^\circ$. 
        \item 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 $F$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): For each $x \in E$, 
    \[
    \bracs{x}^\circ = \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1}
    \]

    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 \autoref{proposition:polar-gymnastics},
    \[
    A^\circ = \bigcap_{\substack{\alpha \in K \\ |\alpha| \ge 1}}\alpha A^\circ
    \]

    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. 
\end{proof}

\begin{proposition}
\label{proposition:equicontinuous-polar}
    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
    \begin{enumerate}
        \item $A$ is equicontinuous.
        \item $A^\circ \in \cn_E(0)$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    By \autoref{proposition:equicontinuous-linear}, $A$ is equicontinuous if and only if
    \[
    \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)
    \]

    (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 $\bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)$. 
\end{proof}


\begin{theorem}[Bipolar Theorem]
\label{theorem:bipolar}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. For each $A \subset F$,
    \[
    A^{\circ\circ} = \ol{\conv}(A \cup \bracs{0})
    \]

    with respect to the $\sigma(E, F)$-topology. 
\end{theorem}
\begin{proof}[Proof, {{\cite[IV.1.5]{SchaeferWolff}}}. ]
    By \autoref{proposition:polar-properties}, $A^{\circ \circ}$ 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 \hypreref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
    \begin{enumerate}
        \item $\phi$ is $\sigma(E, F)$-continuous.
        \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$. 
    \end{enumerate}

    If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by \autoref{proposition:polarisation-linear}, 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 \autoref{lemma:duality-dual}, there exists $y \in F$ such that $\dpn{x, y}{\lambda} = \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,
    \[
    \text{Re}\dpn{x, y}{\lambda} = \text{Re}\Phi(x) = \phi(x)
    \]

    Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda} > 1$, so $x_0 \not\in A^{\circ\circ}$.
\end{proof}

\begin{proposition}
\label{proposition:polar-equicontinuous}
    Let $E$ be a locally convex space, then:
    \begin{enumerate}
        \item For each equicontinuous family $\cf \subset E^*$ equicontinuous, $\cf^\circ \in \cn_E(0)$. 
        \item For each $U \in \cn_E(0)$, $U^\circ$ is equicontinuous. 
    \end{enumerate}
\end{proposition}




