From 225794ff81f449faee811dab1a5e032f1fe15790 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 28 Apr 2026 18:07:27 -0400 Subject: [PATCH] Fixed typo in the definition of regulated maps. --- src/fa/rs/regulated.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/fa/rs/regulated.tex b/src/fa/rs/regulated.tex index 6f1bd2a..624d8c1 100644 --- a/src/fa/rs/regulated.tex +++ b/src/fa/rs/regulated.tex @@ -37,7 +37,7 @@ \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 $G: [a, b] \to F$ 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} \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 $E$.