Is the convex envelope always equal to a convex combination of the original function?

Question

Let $F:[a,b) \to [0,\infty)$ be a $C^1$ function, and let $\hat F$ be the (lower) convex envelope of $F$, i.e. $$ \hat F(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[a, b)$}\,,\, h \le F \} \, . $$

Let $c \in (a,b)$. Do there exist $x,y \in [a,b)$ and $\lambda \in [0,1]$ such that $c = \lambda \, x + (1-\lambda)\, y$ and $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$?

We always have $$ \hat F(c) \le \lambda \, \hat F(x) + (1-\lambda) \, \hat F(y) \le \lambda \, F(x) + (1-\lambda) \, F(y), $$ so $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$ implies that $\hat F(x)=F(x), \hat F(y)=F(y)$.

Here is an attempt at a proof:

Suppose that $\hat F(c)<F(c)$. Define $$ x=\sup \{ t<c \, | \, \hat F(t)=F(t)\}, y=\inf \{ t>c \, | \, \hat F(t)=F(t)\}. $$ Then $x<c<y$, and $\hat F(x)=F(x), \hat F(y)=F(y)$.

My guess is that $\hat F$ should be affine on $[x,y]$, which implies the claim.

I think that this can be proved by assuming by contradiction...but I am having trouble completing the proof.

Noe that $F$ is $C^1$ implies that $\hat F$ is $C^1$, and we also have $F'(x)=\hat F'(x), F'(y)=\hat F'(y)$.

Comment:

I actually don't think that $F \in C^1$ is necessary here. I think that $F$ being continuous should be sufficient.

Answer 1

A counterexample is $F:[a, b) \to [0, \infty)$, $F(x) = (x-a)^2(b-x)^2$, with $\hat F(x) = 0$.

Answer 2

Isn't $\hat F$ determined by the property that its epigraph is the convex hull of the epigraph of $F$? If so the result should follow pretty quickly.

Answer 3

$\newcommand{\epi}{\operatorname{epi}}$

Using Rahul's suggestion, we prove that the answer is positive, if $F$ is defined and continuous on the closed interval $[a,b]$. I think that the same proof holds when the domain is $[a,\infty)$.

It is known that $\epi \hat F =\overline{{ \operatorname{conv}( \epi F})},$ where $\epi F$ is the epigraph of $F$. Furthermore, $\epi F$ is closed if and only if $F$ is lowersemicontinuous. (Here we use the fact that the domain is closed, since otherwise $F$ can be continuous but its epigraph won't be closed, since limit points at the end would be missing).

In particular, if $F$ is continuous, then $\epi F $ is closed. It can be proved that ${ \operatorname{conv} (\epi F})$ is also closed, thus $\epi \hat F = \operatorname{conv}( \epi F)$. This implies that $ (c,\hat F(c))\in \operatorname{conv}( \epi F). $

Carathéodory theorem implies that $(c,\hat F(c)) $ is a convex combination of at most $3$ points from $ \epi F$. Since $ \epi F$ is connected, a sharpened version of Carathéodory's theorem implies that taking convex combinations of two points suffice, i.e.

$(c,\hat F(c))=\lambda (x,r)+(1-\lambda) (y,s)$, where $(x,r),(y,s) \in \epi F$ or $r \ge F(x),s \ge F(y)$. This implies that

$$ \lambda F(x)+(1-\lambda) F(y) \le \lambda r+(1-\lambda) s =\hat F(c). $$ Since we already have the reversed inequality $$ \hat F(c) \le \lambda F(x)+(1-\lambda) F(y) $$ the result follows.

Edit:

It does not work on unbounded domains. Take $F(x)=1-\frac{1}{x+1}$ on $[0,\infty)$. Then it is not hard to see that $\operatorname{conv} (\text{epi} F)=\{(0,0)\} \cup [0,\infty) \times (0,\infty)$ is not closed. In that case $\hat F=0$, and $\epi \hat F =\overline{{ \operatorname{conv}( \epi F})}=[0,\infty) \times [0,\infty)$.