A convex function can have a nonconvex maximum set

The interval $[-1,1]$ is convex since $-1\le s,t\le 1$ implies $-1\le (1-r)s+rt\le 1$ for $0\le r\le 1$. It is closed, its complement being the open rays $(-\mathrm{\infty },-1)$ and $(1,\mathrm{\infty })$, and bounded by one in absolute value. Hence it is nonempty compact convex. For $s,t\in K$, $\mathrm{\mid }q(s)-q(t)\mathrm{\mid }=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }s+t\mathrm{\mid }\le 2\mathrm{\mid }s-t\mathrm{\mid }$, which proves continuity on $K$ directly.

For $s,t\in K$ and $0\le r\le 1$, $(1-r)s^2+r{t}^2-((1-r)s+rt{)}^2=r(1-r)(s-t{)}^2\ge 0.$ Thus $q((1-r)s+rt)\le (1-r)q(s)+rq(t)$, the defining convexity inequality for a function. It includes $r=0,1$ and $s=t$, where equality holds.

On $K$, $q(t)\le 1$ with equality exactly when $t=-1$ or $t=1$, because $1-t^2=(1-t)(1+t)$ and both factors are nonnegative. The maximum set is therefore $M=\{-1,1\}$. Its midpoint is zero, and $q(0)=0<1$, so $0\mathrm{\notin }M$. Thus $M$ is not convex and cannot be a face. This refutes the claim with a continuous convex function on a compact convex set.

Nevertheless $M$ is extremal. If $(1-r)s+rt=1$ with $s,t\in K$ and $0<r<1$, then $(1-r)(1-s)+r(1-t)=0$. Each summand is nonnegative and each coefficient is positive, so $s=t=1$. Similarly a combination equal to $-1$ gives $(1-r)(s+1)+r(t+1)=0$ and forces $s=t=-1$. Thus the endpoint property holds even though convexity fails. All witnesses and computations are explicit and choice-free.