Proposition 16.2.5.label 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
- (1)
$A$ is equicontinuous.
- (2)
$A^{\circ} \in \cn_{E}(0)$.
Proof. By Proposition 10.13.1, $A$ 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 $\bigcap_{\phi \in A}\phi^{-1}(B_{K}(0, 1)) \in \cn_{E}(0)$.$\square$