How to prove an asymptotic behaviour of this quantitative convexity problem? Let $\lambda_n \in [0,1]$, and let $a_n \in (0,\frac{1}{4}),b_n \in [\frac{1}{4},\infty)$ satisfy the relation that the convex combination $$\lambda_n a_n+(1-\lambda_n)b_n=c > \frac{1}{4} \tag{1}$$ is a constant which does not depend on $n$.
Define  $F:[0, \infty) \to \mathbb R$ by
$$F(s) :=
\begin{cases}
1-2s, & \text{ if  }\, s \le \frac{1}{4} \\
2(\sqrt{s}-1)^2,  & \text{ if  }\, s \ge \frac{1}{4}
\end{cases}
$$
$F \in C^1$ is convex as its derivative $$F'(s)=\begin{cases}
 -2, & s\le\frac{1}{4}, \\
 2\left(1-\frac1{\sqrt{s}}\right), & s\geq\frac14
\end{cases}$$
is non-decreasing.
Now, suppose that
$$
D_n:=F\big(\lambda_n a_n +(1-\lambda_n)b_n\big)-\lambda_nF(a_n)-(1-\lambda_n)F(b_n) \to 0
$$
when $n \to \infty$.

Question: I is true that $\lambda_n \to 0$? 

There are two steps which needs to be done:


*

*Moving the estimate on the convexity gap into the strictly convex region $[1/4,\infty)$. 

*Deducing that every convergent subsequence of $\lambda_n$ converges to $0$ or to $1$. 
(since $\lambda_n(1-\lambda_n)$ should be small-see strongly convex functions).
The constraint $(1)$ can be satisfied either when $\lambda_n \to 0$ or when $\lambda_n \to 1$. In the latter case, we must have $b_n \to \infty$.
This should imply that must have passed through a large region where $f$ is strongly convex, thus   the convexity gap $D_n$ is large, which is a contradiction. 
I think that there should be a (relatively) simple conceptual argument based on strong convexity properties of $F$ that establishes that. Note that $F$ is not twice differentiable at $\frac{1}{4}$, and that $\lim_{x \to \infty} F''(x)=0$, so $F$ becomes less convex when we get far away.
 A: Let $\alpha_n\in (0,1)$ be such that $\alpha_n \frac{1}{4} + (1-\alpha_n) b_n = c$.
We have $\alpha_n >\lambda_n$ and from convexity we have
$$\lambda_n F(a_n) + (1-\lambda_n) F(b_n) \ge \alpha_n F(\frac{1}{4}) + (1-\alpha_n) F(b_n) \ge F(c).$$
It follows that the assumption $D_n\to 0$ implies that  $\alpha_n F(\frac{1}{4}) + (1-\alpha_n) F(b_n) \to F(c)$. Since $\alpha_n > \lambda_n$, it is enough to prove that $\alpha_n \to 0$.
We first show that $b_n$ is bounded. Indeed, for $x>c$ let $\alpha(x) = \frac{x-c}{x-\frac{1}{4}}$, that is, $\alpha_n = \alpha(b_n)$.
Define
$$g(x) = \alpha F(\frac{1}{4}) + (1-\alpha)F(x).$$
The strict convexity of $F$ implies that $g$ is an increasing function of $x$, as can be seen from a quick drawing or by differentiating $g$ and using the fact that the strict convexity implies that
$$F'(x) > \frac{F(x)-F(a)}{x-a}$$
for $x>a$.
The fact that $g$ is increasing implies that $b_n$ must be bounded.
Once we have that $b_n$ is bounded, by compactness we can assume that $b_n\to b$, $\alpha_n \to \alpha$ and
$$\alpha F(\frac{1}{4}) + (1-\alpha) F(b) \ge F(c) = F(\alpha \frac{1}{4} + (1-\alpha) b).$$
From the strict convexity of $F$ in $[1/4,\infty)$ and the fact that $c>1/4$, it follows that $\alpha = 0$ and $b=c$.
