Criteria for a Subset A to be σ(E, F)-Weakly Bounded
 Proposition 1: Let $(E, F)$ be a dual pair and let $A \subseteq E$. The following statements are equivalent: (1) $A$ is a $\sigma(E, F)$-weakly bounded subset of $E$. (2) The function $p'$ defined of $F$ for all $y \in F$ by $p'(y) := \sup \{ |\langle x, y \rangle| : x \in A \}$ is a seminorm on $F$. (3) The polar $A^{\circ}$ (in $F$) is an absorbent subset of $F$.
• Proof: $(1) \Rightarrow (2)$ Suppose that $A$ is a $\sigma(E, F)$-bounded subset of $E$.
• Then, for all $y \in F$, we have that $\langle A, y \rangle$ is a bounded set of nonnegative real numbers. Hence, $\sup \{ |\langle x, y \rangle | : x \in A \}$ is not infinite for any $y \in F$, so that $p' : F \to [0, \infty)$. Then it is easy to check that $p'$ satisfies the three properties to be a seminorm on $F$.
• $(2) \Rightarrow (3)$ Suppose that $p'$ is a seminorm on $F$. Let $y \in F$. Then $p'(y) = \sup \{ |\langle x, y \rangle | : x \in A \}$ is bounded. So there exists an $\lambda > 0$ such that $|\langle x, y \rangle| \leq \lambda$ for all $x \in A$. Then observe that $y \in \lambda A^{\circ}$, for $\lambda A^{\circ}$ is the set of all $z \in F$ for which $\sup \{ |\langle x, z \rangle| : x in A \} \leq M$, and we have that $\sup \{ |\langle x, y \rangle| : x \in A \} \leq M$ since $|\langle x, y \rangle| \leq M$ for all $x \in A$.
• Since $A^{\circ}$ is absolutely convex (see one of the propositions on The Polar of a Set page), we have by one of the propositions on the Absolutely Convex Sets of Vectors page that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $\lambda A^{\circ} \subseteq \mu A^{\circ}$. Thus $y \in \mu A^{\circ}$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq \lambda$, so $A^{\circ}$ is absorbent.
• $(3) \Rightarrow (1)$ Suppose that $A^{\circ}$ is absorbent. Then, for each $y \in F$, there exists a $\lambda > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $y \in \mu A^{\circ}$. In particular, $y \in \lambda A^{\circ}$. Therefore $\sup \{|\langle x, y \rangle| : x \in A \} \leq \lambda$ for all $x \in A$, and so $|\langle x, y \rangle| \leq \lambda$ for all $x \in A$. Thus $\langle x, y \rangle$ is bounded for each $y$, and so $A$ is $\sigma(E, F)$-weakly bounded. $\blacksquare$