Does this strong convexity estimate hold? Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function, and let $r_0<r_1$ be positive fixed constants. Let $$a<r_0<r_1<c<b, \tag{1}$$ and let $\lambda \in [0,1]$ satisfy
$ \lambda a +(1-\lambda)b=c. $
Set $D(a,b,c)=\lambda F(a)+(1-\lambda)F(b)-F(c) $.

Question: Does there exist a constant $m>0$ (which may depend on $f,r_0,r_1$ but not on $a,b,c$) such that $
D(a,b,c) \ge m\lambda(1-\lambda)(r_1-r_0)^2
$ for any choice of $a,b,c$ satisfying condition $(1)$?

Here is the key point:
If $f'' \ge m$, then $f$ is strongly convex with parameter $m$, so
$$
D(a,b,c) \ge \frac{1}{2}m\lambda(1-\lambda)(b-a)^2 \ge \frac{1}{2}m\lambda(1-\lambda)(r_1-r_0)^2 \tag{2}
$$
as required. However, in our case, $c$ and $b$ can be arbitrarily large, and $F$ can become "less convex" (closer to being affine) when $x \to \infty$. In other words, if $\lim_{x \to \infty}F''(x)=0$, then the lower bound $(2)$ becomes the trivial bound
$$
D(a,b,c) \ge \frac{1}{2} (\inf F'')\lambda(1-\lambda)(b-a)^2=0.
$$
So, "naive application" of strong convexity does not apply here as is. However, my intuition is that even if $\lim_{x \to \infty}F''(x)=0$, we should somehow encounter "the strong convexity content" which lies between the fixed $r_0$ and $r_1$ so the "convexity gap" $D(a,b,c)$ should be bounded away from zero.
I thought to express $D(a,b,c)$ as some integral of $F''$ over a domain which contains $[r_0,r_1]$ but so far without success.
 A: If suffices to require that $F$ is strictly convex and differentiable on an interval $I \subset \Bbb R$. (Even the differentiability requirement can be dropped, see the remarks at the end of the answer.)
For $a, b \in I$ with $a < b$ and $c = \lambda a + (1 - \lambda) b$ with $0  \le \lambda \le 1$ we can write
$$
 D(a, b, c) = \lambda F(a)+(1-\lambda)F(b)-F(c) \\
 = \lambda \bigl \{ F(a) - F(c) - (a-c)F'(c) \bigr\}
 + (1- \lambda) \bigl \{F(b) - F(c) - (b-c)F'(c)\bigr\} \, .
$$
This suggests to introduce
$$
 H(u, v) = F(u) - F(v) - (u-v) F'(v)
$$
for $u, v \in I$. $H$ has the following properties:

*

*$H(u, v) > 0$ if $u \ne v$.

*$H(u_1, v) > H(u_2, v)$ if $u_1 < u_2 \le v$, i.e. $H(u,v)$ is decreasing in $u$ as long as $u \le v$.

*$H(u, v_1) < H(u, v_2)$ if $u \le v_1 < v_2$, i.e. $H(u, v)$ is increasing in $v$ as long as $u \le v$.

Property (1) is a direct consequence of the strict convexity: $F(u)$ is larger than the corresponding value of the tangent line at $x=v$.
For property (2) we assume $u_1 < u_2 \le v$ and compute
$$
 H(u_1, v) - H(u_2, v) = F(u_1) - F(u_2) - (u_1 - u_2) F'(v) \\
\ge  F(u_1) - F(u_2) - (u_1 - u_2) F'(u_2) 
= H(u_1, u_2) > 0 \, .
$$
Here we used that $F'$ is increasing.
For property (3) we assume $u \le v_1 < v_2$ and compute
$$
 H(u,v_1) - H(u, v_2) = -F(v_1) - (u-v_1)F'(v_1) + F(v_2) + (u-v_2) F'(v_2) \\
\le -F(v_1) - (u-v_1)F'(v_2) + F(v_2) + (u-v_2) F'(v_2) \\
= -H(v_1, v_2) < 0 \, .
$$
With these tools, estimating $D(a, b, c)$ from below becomes easy. If  $a \le r_0 < r_1 \le c < b$ then
$$
 D(a, b, c) = \lambda H(a, c) + (1-\lambda)H(b,c) \\
\ge \lambda H(a, c) \ge \lambda H(r_0, r_1) 
\ge \lambda(1- \lambda) H(r_0, r_1) \\
= m \lambda(1-\lambda)  (r_1-r_0)^2
$$
with $m$ defined as
$$ 
m = \frac{H(r_0, r_1)}{(r_1-r_0)^2}  = \frac{F(r_0) - F(r_1) - (r_0 - r_1) F'(r_1)}{(r_1-r_0)^2} > 0 \, .
$$
Remarks:

*

*The assumption that $F$ is only defined on $[0, \infty)$ with values in $[0, \infty)$ was not used in the proof.

*The differentiability requirement can also be dropped. A convex function has one-sided derivatives in every inner point of the interval. The above proof still works if we replace $F'$ by the right (or left) derivative.

