Is the convex combination realizing the convex envelope unique? 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  $[x,y]$.

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
$$ \hat F(x) = F(x)=1-x,\quad x = x_1,x_2,y_1,y_2$$
