Corollary 17.3.4.label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\topo$ be a topology on $E$ consistent with $\dpn{E, F}{\lambda}$, and $B \subset E$, then $B$ is bounded with respect to $\topo$ if and only if $B$ is bounded with respect to $\sigma(E, F)$.

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,

\[A = \bracsn{\phi \in F|\ |\dpn{x, \phi}{\lambda}| \le 1 \forall x \in B}\]

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 barreled by Proposition 12.4.4. By continuity of the inclusion map, $A \cap E_{B}$ is a barrel in $E_{B}$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_{B} \subset \lambda A$.$\square$

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