85 lines
5.7 KiB
TeX
85 lines
5.7 KiB
TeX
\section{The Mackey-Arens Theorem}
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\label{section:mackey-arens}
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\begin{definition}[Consistent]
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\label{definition:consistent-lct-dual}
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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 $(E, \mathcal{T})^* = F$.
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\end{definition}
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\begin{lemma}
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\label{lemma:consistent-same-convex}
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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.
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\end{lemma}
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\begin{proof}
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By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}.
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\end{proof}
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\begin{theorem}[Mackey-Arens]
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\label{theorem:mackey-arens}
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Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$.
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\begin{enumerate}
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\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 $F$.
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\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 $\dpn{E, F}{\lambda}$.
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\end{enumerate}
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Moreover,
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\begin{enumerate}
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\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 $\dpn{E, F}{\lambda}$.
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\end{enumerate}
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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 $\tau(E, F)$.
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\end{theorem}
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\begin{proof}
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(E): Let $\cf \subset F$ be a $\mathcal{T}$-equicontinuous subset, then $\cf^\circ \in \cn_{\mathcal{T}}(0)$. If $\cf$ is circled, then
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\[
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\cf^\circ = \bracsn{x \in E|\ |\dpn{x, y}{\lambda}| \le 1 \forall y \in \cf}
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\]
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Thus $\mathcal{T}$ is finer than the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$.
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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
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\[
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\fB = \bracsn{\cf^\circ| \cf \subset F, \cf \text{ is } \mathcal{T}\text{-equicontinuous}}
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\]
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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 $F$.
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(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 $F = E^*$.
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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
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\[
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\bracsn{x \in E|\dpn{x, \psi}{E} \le \mu \forall \psi \in A} \subset \bracsn{x \in E|\dpn{x, \phi}{E} \le 1} = \bracs{\phi}^\circ
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\]
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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
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\[
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A^\circ = \bracsn{x \in E|\dpn{x, \psi}{E} \le 1 \forall \psi \in A} \subset \bracs{\phi}^\circ
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\]
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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}, $A^{\circ\circ} \supset \bracs{\phi}^{\circ\circ}$.
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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}, $\phi \in A^{\circ\circ} = A \subset F$.
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(M): By (C), the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets is consistent with $\dpn{E, F}{\lambda}$.
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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.
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\end{proof}
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\begin{definition}[Mackey Space]
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\label{definition:mackey-space}
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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 $\dpn{E, E^*}{E}$.
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\end{definition}
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\begin{proposition}
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\label{proposition:barreled-mackey}
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Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space.
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\end{proposition}
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\begin{proof}
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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.
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\end{proof}
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