How bad can be the set where a convex and a continuous function agree?

Question

Let $F:[a,b] \to \mathbb R$ be a continuous function, and let $G[a,b] \to \mathbb R$ be a convex function. Define $A=\{ x \in [a,b] \, | \, F(x)=G(x)\}$, and suppose that $A$ is not empty.

Must $A$ be a union (finite or countable) of pairwise disjoint closed intervals?(Or can it be something more pathological, like the Cantor set?)

Does the answer change if we assume in addition that $F$ is monotone?

Answer 1

The answer to both questions: $A$ can be any closed subset of $[a,b].$ (Note that such sets can be a bit crazy, for example $A$ could be a closed set of positive measure with no interior.)

Proof: Let $A\subset [a,b]$ be closed. Then there exists a $C^\infty$ function $F$ on $[a,b]$ whose zero set is precisely $A.$ Take $G\equiv 0.$ Then $G$ is convex and the set where $F=G$ is $A.$

Getting fancier, we can choose $c>0$ such that

$$c(|F'(x)|+|F''(x)|)\le 1/2,\,x\in [a,b].$$

Then the functions $cF(x)+x+(x-a)^2,$ $x+(x-a)^2$ are $C^\infty,$ strictly increasing, strictly convex, and agree precisely on the set $A.$

Answer 2

If $0<a<b<\frac1 2$ then $x+(x-a)(x-b)$ is a continuous increasing function on $[a,b]$ such that $f(x)=x$ iff $x=a$ or $x=b$. No apply this to each of the intervals $(\frac 1 {n+1}, \frac 1 n)$ with $n >2$. Let $G(x)=x$ for all $x$. Then $F(x)=G(x)$ iff $x =\frac 1n$ for some $n >2$. Hence your set is not a union of intervals. Clearly $G$ is convex and $F$ is an increasing continuous function on $[0,\frac 1 2]$.