12.14 Equicontinuous Families of Linear Maps
Proposition 12.14.1.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, [IV.4.2, SW99]. (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 12.14.2.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 12.14.3 ([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 12.13.8, $\alg' \subset \hom(E; F)$. By Theorem 6.5.4, $\alg'$ is equicontinuous.$\square$
Theorem 12.14.4 (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 12.14.1.
(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 12.14.1.
(E1) $\Rightarrow$ (C1) + (C2): By the Arzelà-Ascoli Theorem and Proposition 12.14.3.$\square$
Lemma 12.14.5.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 12.14.6 (Banach-Alaoglu).label Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^{*}$ be equicontinuous, then $\alg$ is relatively compact with respect to $\sigma(E^{*}, E)$.
Proof. For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is relatively compact by Proposition 12.14.2. By the Arzelà-Ascoli Theorem,
- (C2)
The $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is equicontinuous.
- (C3)
The $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is compact.
By Proposition 12.13.8, the $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}\ol{\alg(x)}$ is a subset of $\hom(E; K)$. Hence the $\sigma(E^{*}, E)$-closure of $\alg$ in $E^{*}$ is compact.$\square$
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