Added remark.

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}
+
+