10.13 Equicontinuous Families of Linear Maps
Proposition 10.13.1 ([IV.4.2, SW99]).label Let $E, F$ be TVSs over $K \in RC$ and $\alg \subset \hom(E; F)$, then the following are equivalent:
- (1)
$\alg$ is uniformly equicontinuous.
- (2)
$\alg$ is equicontinuous.
- (3)
$\alg$ is equicontinuous at $0$.
- (4)
For each $V \in \cn_{F}(0)$, there exists $U \in \cn^{o}(E)$ such that $\bigcup_{T \in \alg}T(U) \subset V$.
- (5)
For each $V \in \cn_{F}(0)$, $\bigcap_{T \in \alg}T^{-1}(V) \in \cn_{E}(0)$.
Proof. (5) $\Rightarrow$ (1): Let $V \in \cn_{F}(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_{E}(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$.$\square$
Proposition 10.13.2.label Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_{\sigma}(E; F)$.
Proof. Let $S \in \sigma$ and $U \in \cn_{F}(0)$, then there exists $V \in \cn_{E}(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_{\sigma}(E; F)$.$\square$
Proposition 10.13.3 ([IV.4.3, SW99]).label Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^{E}$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$.
Proof. By Proposition 10.12.8, $\alg' \subset \hom(E; F)$. By Theorem 6.5.4, $\alg'$ is equicontinuous.$\square$
Theorem 10.13.4 (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:
- (B)
$E$ is a Baire space.
- (B’)
$E$ is barrelled and $F$ is locally convex.
and that
- (E2)
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]. (B) + (E2) $\Rightarrow$ (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 (E2), $U$ 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 10.13.1.
(B’) + (E2) $\Rightarrow$ (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 (E2), $U$ is absorbing, and hence a barrel in $E$. By (B’), $U \in \cn_{E}(0)$, $\alg$ is equicontinuous by Proposition 10.13.1.
(E1) $\Rightarrow$ (C1) + (C2): By the Arzelà-Ascoli Theorem and Proposition 10.13.3.$\square$
Lemma 10.13.5.label Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg \subset L^{2}(E, F; G)$ be continuous bilinear maps, then the following are equivalent:
- (1)
$\alg$ is equicontinuous.
- (2)
$\alg$ is equicontinuous at $0$.
Proof. (2) $\Rightarrow$ (1): For each $(x_{0}, y_{0}), (x, y) \in E \times F$ and $\lambda \in \alg$,
For each $U \in \cn_{G}(0)$ circled, there exists circled neighbourhoods $V \in \cn_{E}(0)$ and $W \in \cn_{F}(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_{0} \in \mu W$ and $x_{0} \in \mu V$. Thus if $(x, y) - (x_{0}, y_{0}) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,
and $\lambda(x_{0}, y - y_{0}) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_{0}, y_{0})$.$\square$
Theorem 10.13.6.label Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
- (B)
$E$ is Baire.
- (B’)
$E$ is barrelled and $G$ is locally convex.
and that
- (M)
$E$ and $F$ are both metrisable.
- (E)
For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg}\subset L(F; G)$ is equicontinuous.
then $\alg$ is equicontinuous.
Proof, [III.5.1, SW99]. Let $\seq{(x_n, y_n)}\subset E \times F$ and $\seq{\lambda_n}\subset \alg$ such that $(x_{n}, y_{n}) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and Proposition 6.5.2. By (B) or (B’) and the Banach-Steinhaus Theorem, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_{n}(x_{n}, y_{n}) \to 0$ as $n \to \infty$ by Proposition 6.5.2. By (M) and Proposition 6.5.2, $\alg$ is equicontinuous at $0$, and hence equicontinuous by Lemma 10.13.5.$\square$