\section{The Mackey-Arens Theorem}
\label{section:mackey-arens}

\begin{definition}[Consistent]
\label{definition:consistent-lct-dual}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $\mathcal{T} \subset 2^E$ be a locally convex topology, then $\mathcal{T}$ is \textbf{consistent} with $\dpn{E, F}{\lambda}$ if $(E, \mathcal{T})^* = F$. 
\end{definition}

\begin{lemma}
\label{lemma:consistent-same-convex}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$, then for any $A \subset E$ convex, the $\mathcal{T}$-closure of $A$ and the $\sigma(E, F)$-closure of $A$ coincide. 
\end{lemma}
\begin{proof}
    By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}. 
\end{proof}


\begin{theorem}[Mackey-Arens]
\label{theorem:mackey-arens}
    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. 
    \begin{enumerate}
        \item[(E)] For any locally convex topology $\mathcal{T} \subset 2^E$ consistent with $\dpn{E, F}{\lambda}$, $\mathcal{T}$ is the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 
        \item[(C)] For any saturated covering ideal $\sigma \subset 2^{F}$ consisting of relatively $\sigma(F, E)$-compact sets, the $\sigma$-uniform topology on $E$ is consistent with $\dpn{E, F}{\lambda}$. 
    \end{enumerate}

    Moreover, 
    \begin{enumerate}
        \item[(M)] The topology of uniform convergence on relatively $\sigma(F, E)$-compact, convex, and circled sets is the finest locally convex topology on $E$ consistent with $\dpn{E, F}{\lambda}$. 
    \end{enumerate}

    and strongest locally convex topology on $E$ consistent with $\dpn{E, F}{\lambda}$ is the \textbf{Mackey topology} of $\dpn{E, F}{\lambda}$ on $E$, denoted $\tau(E, F)$. 
\end{theorem}
\begin{proof}
    (E): Let $\cf \subset F$ be a $\mathcal{T}$-equicontinuous subset, then $\cf^\circ \in \cn_{\mathcal{T}}(0)$. If $\cf$ is circled, then
    \[
    \cf^\circ = \bracsn{x \in E|\ |\dpn{x, y}{\lambda}| \le 1 \forall y \in \cf}
    \]

    Thus $\mathcal{T}$ is finer than the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 
    
    Conversely, for any convex and closed neighbourhood $U \in \cn_{\mathcal{T}}(0)$, $U^\circ$ is equicontinuous. By the \hyperref[Bipolar Theorem]{theorem:bipolar}, $U^{\circ \circ} = U$, so the family
    \[
    \fB = \bracsn{\cf^\circ| \cf \subset F, \cf \text{ is } \mathcal{T}\text{-equicontinuous}}
    \]

    is a fundamental system of neighbourhoods at $0$ for $\mathcal{T}$. Hence $\mathcal{T}$ is coarser than the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 

    (C): Let $\mathcal{T}$ be the $\sigma$-uniform topology on $E$, and $E^*$ be the dual of $(E, \mathcal{T})$. Since $\sigma$ is covering, it contains all singletons, so $\mathcal{T} \supset \sigma(E, F)$, and $F \subset E^*$. Thus it is sufficient to show that $F = E^*$. 

    Let $\phi \in E^*$ and $\bracs{\phi}^\circ$ be the polar of $\phi$ with respect to $\dpn{E, E^*}{E}$. Since $\phi$ is continuous, $\bracs{\phi}^\circ \in \cn_{\mathcal{T}}(0)$. Thus there exists $A \in \sigma$ and $\mu > 0$ such that
    \[
    \bracsn{x \in E|\dpn{x, \psi}{E} \le \mu \forall \psi \in A} \subset \bracsn{x \in E|\dpn{x, \phi}{E} \le 1} = \bracs{\phi}^\circ
    \]

    Since $\sigma$ is covering and saturated, assume without loss of generality that $\mu = 1$ and that $A$ is convex, circled, and $\sigma(F, E)$-compact with $0 \in A$. In which case, let $A^\circ$ be the polar of $A$ with respect to $\dpn{E, E^*}{E}$, then
    \[
    A^\circ = \bracsn{x \in E|\dpn{x, \psi}{E} \le 1 \forall \psi \in A} \subset \bracs{\phi}^\circ
    \]

    Let $A^{\circ\circ}$ and $\bracs{\phi}^{\circ\circ}$ be the bipolar of $A$ and $\bracs{\phi}$ with respect to $\dpn{E, E^*}{E}$, then by \autoref{proposition:polar-gymnastics}, $A^{\circ\circ} \supset \bracs{\phi}^{\circ\circ}$. 

    Now, $A$ is $\sigma(F, E)$-compact, and hence $\sigma(E^*, E)$-compact in $E^*$\footnote{Closedness is insufficient because $F$ is $\sigma(E^*, E)$-dense in $E^*$ by \autoref{lemma:duality-dense}. } by \autoref{proposition:compact-closed}. Thus by the \hyperref[Bipolar Theorem]{theorem:bipolar}, $\phi \in A^{\circ\circ} = A \subset F$. 

    (M): By (C), the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets is consistent with $\dpn{E, F}{\lambda}$. 
    
    On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
\end{proof}

\begin{definition}[Mackey Space]
\label{definition:mackey-space}
    Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 
\end{definition}

\begin{proposition}
\label{proposition:barreled-mackey}
    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
\end{proposition}
\begin{proof}
    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
\end{proof}




