18.3 The Mackey-Arens Theorem

Proof. By the Hahn-Banach theorem.$\square$

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

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 Bipolar Theorem, $U^{\circ \circ}= U$, so the family

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

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

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 Proposition 18.2.3, $A^{\circ\circ}\supset \bracs{\phi}^{\circ\circ}$.

Now, $A$ is $\sigma(F, E)$-compact, and hence $\sigma(E^{*}, E)$-compact in $E^{*}$^[1] by Proposition 5.16.4. Thus by the Bipolar Theorem, $\phi \in A^{\circ\circ}= A \subset F$.

(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 Banach-Alaoglu Theorem, 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.$\square$

Proof. Assume without loss of generality that $\topo = \tau(E, F)$. Suppose that $B$ is $\sigma(E, F)$-bounded. Let $A \subset F$ be convex, circled, and $\sigma(F, E)$-compact. By continuity of dual pairing,

is a convex, circled, and closed subset. Given that $B$ is $\sigma(E, F)$-bounded, $\sup_{x \in B}|\dpn{x, \phi}{\lambda}| < \infty$ for all $x \in E$. Thus $A$ is absorbing and hence a barrel.

Since $B$ is compact, the auxiliary space $E_{B}$ is a Banach space. In particular, $E_{B}$ is barrelled by Proposition 13.4.4. By continuity of the inclusion map, $A \cap E_{B}$ is a barrel in $E_{B}$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_{B} \subset \lambda A$.$\square$

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 barrelled, $V \in \cn_{E}(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space.$\square$