Let $F:[a,b] \to [0,\infty)$ be a continuous function.
Let $c \in [a,b]$. Given $x\le c \le y$, suppose that $\lambda \in [0,1]$ and $c = \lambda \, x + (1-\lambda)\, y$. ($\lambda$ is unique unless $x=y=c$).
Define $D(x,y)=\lambda \, F(x) + (1-\lambda) \, F(y)$, and set $\hat F(c) =\inf_{x\le c \le y} D(x,y)$.
Question: Is there a "direct" way to prove that $\hat F$ is convex? Is $\hat F$ still convex when $b=\infty$ (the domain is unbounded)?
In the case where our domain $[a,b]$ is compact, $\hat F(c) =\min_{x\le c \le y} D(x,y)$ (since $F$ continuous).
Here is an indirect proof:
Let $CF$ the convex envelope of $F$, i.e. $$ CF(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[a, b)$}\,,\, h \le F \} \, . $$ Since $CF$ is convex, we have $$ CF(c) \le \lambda \, CF(x) + (1-\lambda) \, CF(y) \le \lambda \, F(x) + (1-\lambda) \, F(y)=D(x,y), $$ so $CF(c) \le \hat F(c)$.
However, as proved here, there exist $x,y \in [a,b]$ and $\lambda \in [0,1]$ such that $c = \lambda \, x + (1-\lambda)\, y$ and $CF(c) = \lambda \, F(x) + (1-\lambda) \, F(y)=D(x,y)\ge \hat F(c)$.
Thus, $\hat F=CF$, so $\hat F$ is convex as required. However, the proof that $CF(c)$ can actually be realized as a convex sum used Caratheodory theorem, and the characterizaion of the epigraph of the convex envelope $CF$ in terms of the epigraph of $F$.
Also, this "realization fact" does not hold when $b= \infty$.
Is there a more direct way to prove this, without using the term "convex envelope"? or facts about it?
