Let $F:[0,\infty) \to [0,\infty)$ be a $C^1$ 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 strictly convex for some $\epsilon>0$.
Let $QF$ be the convex envelope of $F$. Does $QF$ have the same monotonicity properties as $F$?
i.e. is $QF$ strictly increasing on $[1,\infty)$, and strictly decreasing on $[0,1]$?
Best Answer
Your conjecture is almost correct. The convex envelope of $F$ is strictly decreasing on the (bounded) interval $[0, 1]$, and strictly increasing or identically zero on the (unbounded) interval $[1, \infty)$. $F$ need not even be continuous for this conclusion, and the strict convexity near $x=1$ is also not needed.
In the following, $\hat F$ denotes the (lower) convex envelope of $F$, that is $$ \hat F(x) = \sup \{ h(x) \mid \text{$h: \operatorname{dom}(F) \to \Bbb R$ is convex}, h \le F \} \, . $$
Then we have the following result for unbounded intervals:
And for bounded intervals:
For a proof of the statement about unbounded intervals we distinguish two cases:
Case 1: $\liminf_{x \to \infty} F(x)/x = 0$. Then also $\liminf_{x \to \infty} G(x)/x = 0$, and that implies that $G$ is identically zero on $[a, \infty)$.
Case 2: $\liminf_{x \to \infty} F(x)/x > 0$. Then $F(x) > cx$ for some constant $c > 0$ and $x \ge x_1 > a$.
For $x_0 \in (a, x_1)$ set $m = \min(c, \frac{f(x_0)}{x_1 - x_0})$ and consider the function $h(x) = m(x-x_0)$. $h$ is convex with $h \le F$, so that $\hat F(x) \ge h(x) > 0$ on $(x_0, \infty)$.
Since $x_0$ can be arbitrarily close to $a$ it follows $\hat F(x) > 0$ on $(a, \infty)$. This implies that $\hat F$ is strictly increasing on $[a, \infty)$.
A similar reasoning as in the second case can be used to prove the statement about bounded intervals.