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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.

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 $x$).

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).

Asaf Shachar
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    Why should the partition be finite? – Michael Sep 04 '20 at 05:26
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    What about this? The set $A { x \mid CF(x) < F(x) }$ is open in $[a, b]$ and therefore a (finite or countable) collection of open intervals $(a_i, b_i)$. On each interval is $CF$ affine, and on the complement of $A$ is $CF = F$. – Martin R Sep 04 '20 at 07:00
  • @MartinR I have thought more on your suggestion. I do think that it may be in the right direction, but it does not settle the question yet. The problem is that the complement can be pathological. I have edited the question to address this. In particular, I don't know how to decide the following subquestion: Let $x \in (a,b)$. Does 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 $F$). I have stated all of this in the body of the question. – Asaf Shachar Sep 09 '20 at 10:52

2 Answers2

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

copper.hat
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  • Oh , I guess you are right. Thanks. But is there an infinite partition? The essential point of the question was whether $CF$ must zig-zag between affine parts and $F$-like parts. I think that I will keep the question open for now, to see if someone can answer the "infinite partition" version of it. (Instead of opening a new one). I hope that's OK with you. – Asaf Shachar Sep 04 '20 at 05:37
  • No problem. I suspect some pretty wild behaviour is possible such as with $x \mapsto -\sqrt{x} \cdot 1_\mathbb{Q}(x)$. – copper.hat Sep 04 '20 at 05:41
  • Thanks. I have updated the question to state more clearly what is the "infinite" version I am expecting for. Regarding your suggestion, I think that $CF(x)=-\sqrt{x}$ (as $CF$ must be continuous). Anyway, I assumed that $F$ itself is continuous, so this might rule out some pathologies. I guess we shall wait and see:) – Asaf Shachar Sep 04 '20 at 05:47
  • What happened? ${}{}$ – copper.hat Sep 10 '20 at 15:07
  • Martin R gave a very interesting counter-example to the "infinite partition" version of the question: https://math.stackexchange.com/a/3820263/104576 – Asaf Shachar Sep 13 '20 at 06:30
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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 $F$.

Martin R
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