Adjusted labeling of conditions for Banach-Steinhaus.
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Bokuan Li
2026-05-30 20:40:38 -04:00
parent edde66facc
commit 47145cdf58
2 changed files with 7 additions and 7 deletions

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@@ -73,8 +73,8 @@
\label{theorem:uniform-boundedness}
Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
\begin{enumerate}
\item[(B)] $E$ is a Banach space.
\item[(E2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
\item[(B1)] $E$ is a Banach space.
\item[(B2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
\end{enumerate}
then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.

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@@ -37,13 +37,13 @@
\label{theorem:banach-steinhaus}
Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
\begin{enumerate}
\item[(B)] $E$ is a Baire space.
\item[(B')] $E$ is barrelled and $F$ is locally convex.
\item[(B1)] $E$ is a Baire space.
\item[(B1')] $E$ is barrelled and $F$ is locally convex.
\end{enumerate}
and that
\begin{enumerate}
\item[(E2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$.
\item[(B2')] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$.
\end{enumerate}
then
@@ -54,14 +54,14 @@
\end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ]
(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,
(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,
\[
W \subset nU - nU = nU + nU = 2nU
\]
so $U \in \cn_E(0)$, and $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}.
(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 \autoref{proposition:equicontinuous-linear}.
(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 \autoref{proposition:equicontinuous-linear}.
(E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}.
\end{proof}