= Functional calculus preserves closed faces of the positive unit interval
{title2=$a\in F\implies g(a)\in F$}
Let $F$ be a nonempty <norm>-closed <face of a convex set> $[0,1_A]$ in a unital <C-star algebra>, with $0\notin F$. Every $a\in F$ has <norm> one: otherwise $a=\|a\|(a/\|a\|)+(1-\|a\|)0$ is a nontrivial <convex> decomposition forcing zero into the face. If $g:[0,1]\to[0,1]$ is continuous and nondecreasing with $g(0)=0,g(1)=1$, then $g(a)\in F$.
To prove this, for $n\geq2$ replace $g$ near the endpoints by a function $g_n$ which is zero on $[0,1/(2n)]$, one on $[1-1/(2n),1]$, agrees with $g$ on $[1/n,1-1/n]$, and interpolates linearly in between. Monotonicity and endpoint continuity give $g_n\to g$ uniformly. With $\delta=1/(2n)$, the function $h_n(t)=(t-\delta g_n(t))/(1-\delta)$ takes values in $[0,1]$: check the two constant regions and use $\delta\leq t\leq1-\delta$ elsewhere. The <continuous functional calculus> gives $a=\delta g_n(a)+(1-\delta)h_n(a)$ with both summands positive contractions. The face property gives $g_n(a)\in F$, and closedness gives $g(a)\in F$.
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