Theorem 17.3.3 (Mackey-Arens).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$.
- (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 $F$.
- (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 $\dpn{E, F}{\lambda}$.
Moreover,
- (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 $\dpn{E, F}{\lambda}$.
and strongest locally convex topology on $E$ consistent with $\dpn{E, F}{\lambda}$ is the Mackey topology of $\dpn{E, F}{\lambda}$ on $E$, denoted $\tau(E, F)$.
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 17.2.3, $A^{\circ\circ}\supset \bracs{\phi}^{\circ\circ}$.
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.$\square$
- Closedness is insufficient because $F$ is $\sigma(E^{*}, E)$-dense in $E^{*}$ by Lemma 17.1.5. keyboard_return
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