From f4e5004e8c2c1457704baba78e96995306926751 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 7 Jul 2026 12:34:05 -0400 Subject: [PATCH] Added remark. --- src/op/c-star/order.tex | 6 ++++++ 1 file changed, 6 insertions(+) diff --git a/src/op/c-star/order.tex b/src/op/c-star/order.tex index 5398e8c..9e812c0 100644 --- a/src/op/c-star/order.tex +++ b/src/op/c-star/order.tex @@ -77,5 +77,11 @@ On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^+) + p(-x^-)$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, $f(x) = f(x^+) + f(-x^-)$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^+(x) = f+(x^+) + f^+(-x^-) = f^+(x^+) = x^+$, and likewise $f^-(x) = x^-$. Therefore the decomposition is given uniquely by the continuous functional calculus. \end{proof} +\begin{remark} +\label{remark:positive-negative-cstar-algebra} + The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. +\end{remark} + +