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Let $F:[0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $[0,1]$. Suppose also that $F|_{(1-\epsilon,1+\epsilon)}$ is convex for some $\epsilon>0$. Suppose that $F$ is not affine on any subinterval.

Let $\hat F(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[0, \infty)$}, h \le F \} \, $ be the convex envelope of $F$. Let $c\in (0,1)$, and suppose that $\hat F(c) < F(c)$.

Question: Let $x,y \in [0,\infty)$ and $\lambda \in [0,1]$ satisfy $c = \lambda \, x + (1-\lambda)\, y$ and $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$. Are such $x,y$ unique?

(Here is an argument for the existence of such $x$ and $y$, under slightly different conditions).

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 $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$ if and only if $\hat F(x)=F(x), \hat F(y)=F(y)$, and $\hat F$ is affine on $[x,y]$.

Asaf Shachar
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  • No, they are not unique. The function can "bump" on, for example, the line $y+x=1$ multiple times – Exodd Aug 30 '20 at 13:48

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Take $c = 1/2$, $$x_1=1/2 -a,\quad y_1=1/2 +a,\\ x_2=1/2 -b, \quad y_2=1/2 +b,$$ where $a<b<1/2$. You can find a function $F$ satisfying all the hypotheses such that $\hat F(c)=1-c< F(c)$, and $$ \hat F(x) = F(x)=1-x,\quad x = x_1,x_2,y_1,y_2$$

Exodd
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