Is the convex envelope of a function $F$ piecewise-affine or piecewise-equals to $F$?

Question

Let $F:[a,b] \to \mathbb R$ be a continuous function, and let $CF$ the convex envelope of $F$, i.e. $$ CF(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[a, b]$}\,,\, h \le F \} \, . $$

Question: Is $CF$ "piecewise-affine or piecewise-equals to $F$"?

More precisely, does there exist a finite partition $(x_i)_{i=1,\dots,n}$, $a=x_1 <x_2 <\dots <x_n=b$, such that the restriction $CF|_{[x_i,x_{i+1}]}$ is either affine or coincides with $F$ for every $i$?

Edit:

As pointed out in this answer, a finite partition does not necessarily exist. Does the answer change if we allow for infinite partitions? i.e. does there exist a sequence $(x_i)_{i \in \mathbb Z}$ satisfying $x_i <x_{i+1}$ and $\cup_{i \in \mathbb Z} (x_i,x_{i+1})=(a,b)$, such that the restriction $CF|_{[x_i,x_{i+1}]}$ is either affine or coincides with $F$ for every $i$?

Here is what I know:

Fact 1: Whenever $c\in (a,b)$ and $CF(c)<F(c)$, there exist $x<y$ such that $c\in (x,y)$, $CF(x)=F(x), CF(y)=F(y)$ and $CF|_{[x,y]}$ is affine.

Indeed, by this answer, there exist $x,y \in [a,b]$ such that $c = \lambda \, x + (1-\lambda)\, y$ and $CF(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$. Since $$ \lambda \, F(x) + (1-\lambda) \, F(y)=CF(c) \le \lambda \, CF(x) + (1-\lambda) \, CF(y) \le \lambda \, F(x) + (1-\lambda) \, F(y), $$ we must have $CF(x)=F(x), CF(y)=F(y)$, and $CF(c)= \lambda \, CF(x) + (1-\lambda) \, CF(y)$. This last equality implies that $CF$ is affine on $[x,y]$, since it's convex.

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Define $A= \{ x \in (a,b) \mid CF(x) < F(x) \}$. $F$ is continuous by our assumption, and $CF$ is continuous since it's convex. Thus $A$ is an open subset of $(a,b)$, so it is a finite or countable union of pairwise disjoint open intervals.

It's not hard to see that $CF$ is affine on each subinterval of $A$ (see below), and on $A^c$, $CF=F$. The problem is that $A^c$ can in principle be pathological-e.g. something like the Cantor set, which is not a countable union of closed intervals.

So, the question is still open. I think a positive achievement would be to prove the following:

Let $x \in (a,b)$. Then there exist $\epsilon>0$ such that $CF|_{[x,x+\epsilon]}$, $CF|_{[x-\epsilon,x]}$ are affine or coincide with $F$. (they don't have to "agree"-one of them may be affine while the other one coincide with $F$). We can assume that $CF(x)=F(x)$. (since otherwise, we know that $CF$ is locally affine around $x$).

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Proof that $CF$ is affine on each subinterval of $A$:

Let $I \subseteq A$ be such an interval. By "fact $1$" $CF$ is locally affine on $I$, i.e. for every $c \in I$ there exist $x,y\in I$ such that $c\in (x,y)$, and $CF|_{[x,y]}$ is affine. Since $I$ is connected, $CF$ must be "globally affine" on $I$ -i.e. it coincides with a single affine function on $I$. (Proof: it's derivative must be locally constant, and a locally constant function on a connected space is globally constant).

Answer 1

A counterexample: Construct $F, G: [-1, 1] \to \Bbb R$ as follows:

• $F(0) = G(0) = 0$.
• $F(\frac 1n) = G(\frac 1n) = \frac 1{n^2}$ for all positive integers $n$.
• $G$ is linear on every interval $[\frac{1}{n+1}, \frac 1n]$.
• $F(x) = G(x) + (x-\frac{1}{n+1})(\frac 1n-x)$ on every interval $[\frac{1}{n+1}, \frac 1n]$, i.e. $F$ is a downward parabola on every interval $[\frac{1}{n+1}, \frac 1n]$.
• $F(x) = F(-x)$ and $G(x) = G(-x)$ for $-1 \le x < 0$, i.e. the functions are extended as even functions to $[-1, 1]$.

Then $G$ is the convex envelope of $F$, but there is no interval $[-\epsilon, 0]$ or $[0, \epsilon]$ on which $G$ is linear or coincides with $F$.

Answer 2

Let $f(x) = \sqrt{x} \sin {1 \over x}$ on $(0,1)$. There is no finite partition.