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Fixed typo in the definition of regulated maps.
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Bokuan Li
2026-04-28 18:07:27 -04:00
parent 6a960a6231
commit 225794ff81
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src/fa/rs/regulated.tex
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@@ -37,7 +37,7 @@
\label{definition:regulated-function} \label{definition:regulated-function}
Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map. Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map.
Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is \textbf{regulated with respect to} $G$ if $f$ is continuous on all discontinuity points of $G$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform norm, then: Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is \textbf{regulated with respect to} $G$ if $G$ is continuous on all discontinuity points of $f$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform norm, then:
\begin{enumerate} \begin{enumerate}
\item Every regulated step map is in $RS([a, b], G)$. \item Every regulated step map is in $RS([a, b], G)$.
\item If $E$ is metrisable, then for any $f \in \text{Reg}([a, b], G; E)$, $f$ is continuous at all but at most countably many points, and $f$ does not share any discontinuity points with $E$. \item If $E$ is metrisable, then for any $f \in \text{Reg}([a, b], G; E)$, $f$ is continuous at all but at most countably many points, and $f$ does not share any discontinuity points with $E$.