Face of a convex set

A convex set $F\subseteq C$ is a face of $C$ if $ty+(1-t)z\in F$, with $y,z\in C$ and $0<t<1$, implies $y,z\in F$. A singleton face is an extreme point. A face of a face is a face of the original convex set.

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 $a=\Vert a\Vert (a/\Vert a\Vert )+(1-\Vert a\Vert )0$ 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 $n\ge 2$ 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 $\delta \le t\le 1-\delta$ 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 $g(a)\in F$.

An exposed face of a convex set $K$ is the nonempty set on which a supporting real linear functional attains its maximum. It is a face of a convex set: if $x=ty+(1-t)z$ with $0<t<1$ attains the maximum and $y,z\in K$, each of $f(y),f(z)$ is at most that maximum, so both equal it. On a complex vector space, take the real part of a complex linear functional. An exposed point is an exposed face consisting of one point.