Theorem 11.14.4 (Banach-Steinhaus).label Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
- (B1)
$E$ is a Baire space.
- (B1’)
$E$ is barrelled and $F$ is locally convex.
and that
- (B2’)
For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$.
then
- (E1)
$\alg$ is equicontinuous.
- (C1)
The product topology and the compact-open topology on $\cf$ coincide.
- (C2)
The closure of $\alg$ in $F^{E}$ is with respect to the product topology is an equicontinuous subset of $L(E; F)$.
Proof, [IV.4.2, SW99]. (B1) + (B2) $\Rightarrow$ (E1): Let $V \in \cn_{F}(0)$ be closed and circled, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is circled and closed. By (B2), $U$ is absorbing, so $E = \bigcup_{n \in \natp}nU$. Since $E$ is Baire, there exists $n \in \natp$, $W \in \cn_{E}(0)$, and $x \in E$ such that $x + W \subset nU$. As $U$ is circled,
so $U \in \cn_{E}(0)$, and $\alg$ is equicontinuous by Proposition 11.14.1.
(B1’) + (B2) $\Rightarrow$ (E1): Let $V \in \cn_{F}(0)$ be convex, circled, and closed, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is convex, circled, and closed. By (B2), $U$ is absorbing, and hence a barrel in $E$. By (B1’), $U \in \cn_{E}(0)$, $\alg$ is equicontinuous by Proposition 11.14.1.
(E1) $\Rightarrow$ (C1) + (C2): By the Arzelà-Ascoli Theorem and Proposition 11.14.3.$\square$
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