Does strict convexity imply asymptotic strict convexity on unbounded domains? Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function, with $F''$ everywhere positive.
Let $\lambda_n \in [0,1],a_n, b_n \in [0,\infty)$ satisfy
$$ \lambda_n a_n +(1-\lambda_n)b_n=c>0 , b_n-a_n \ge \epsilon > 0$$
for some constant $c$, and some $\epsilon >0$.
set $D_n:=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(\lambda_n a_n +(1-\lambda_n)b_n\big) \to 0$, and suppose that $\lim_{n \to \infty}D_n=0$.

Is it true that $\lim_{n \to \infty} \lambda_n(1-\lambda_n)=0$?

(Equivalently, does every convergent subsequence of $ \lambda_n$ converges either to zero or to one?)

If $a_n,b_n$ are bounded sequences, then the answer is positive-since by passing to subsequences we can assume that $a_n \to a,b_n \to b,\lambda_n \to \lambda$, so we get
$$F\big(\lambda a +(1-\lambda)b\big)=\lambda F(a)+(1-\lambda)F(b).
$$
The condition $b_n-a_n \ge \epsilon > 0$ implies that $a \neq b$, so the strict convexity of $F$
implies $\lambda=0$ or $\lambda=1$.

I am not sure how to handle the case where $b_n$ is unbounded (The assumption $b_n-a_n \ge \epsilon$ implies that $a_n \le b_n$, so $a_n \le c$, thus $a_n$ is always bounded).
I guess that even if $b_n$ is unbounded, then since the weighted means of $a_n,b_n$ equal the constant $c$, we must somehow hit the strict convexity on a compact neighbourhood of $c$. (So taking $F$ which becomes "less convex" when $x \to \infty$, where $F''$ decreases to zero limit at infinity, won't produce counter-examples).
 A: The answer is positive.
Since the problem is invariant under the change $\lambda_n \iff 1-\lambda_n$, we may assume that $c=(1-\lambda_n)a_n+\lambda_n b_n$.
First, we note that  $c= a_n +\lambda_n(b_n-a_n)\ge a_n +\lambda_n \epsilon $.
Now, set $\tilde a_n=c-\lambda_n \epsilon \ge a_n \ge 0$
, and $\tilde b_n=c+\epsilon(1-\lambda_n )$.
Then $$c=(1-\lambda_n)\tilde a_n+\lambda_n \tilde b_n=(1-\lambda_n)a_n+\lambda_n b_n$$, and since $\tilde a_n \ge a_n$ it follows that $\tilde b_n \le b_n$.
The convexity of $F$ now implies that 
$$
(1-\lambda_n)F(\tilde a_n)+\lambda_n F(\tilde b_n) \le(1-\lambda_n)F(a_n)+\lambda_n F(b_n),
$$
so $\tilde D_n \le D_n$ where
$$\tilde D_n:=\lambda_nF(\tilde a_n)+(1-\lambda_n)F(\tilde b_n)-F\big((1-\lambda_n) \tilde a_n +\lambda_n\tilde b_n\big).$$
So we replaced $(a_n,b_n)$ with bounded sequences $(\tilde a_n,\tilde b_n)$ satisfying $\tilde b_n-\tilde a_n =\epsilon$, (while keeping the constants $\lambda_n$ the same). This reduces the problem to the case where $b_n$ is bounded, so we are done.

In fact, we can get even a quantitative estimate on how fast the $\lambda_n$ must converge:
Now, $[\tilde a_n,\tilde b_n] \subseteq [\min(c-\epsilon,0),c+\epsilon)$, and $\tilde b_n-\tilde a_n =\epsilon$.
Thus,
$$
D_n \ge \tilde D_n \ge 1/2 (\min_{x \in [\min(c-\epsilon,0),c+\epsilon]} F''(x)) \lambda_n(1-\lambda_n)(\tilde b_n-\tilde a_n)^2=$$ $$1/2 (\min_{x \in [\min(c-\epsilon,0),c+\epsilon]} F''(x)) \epsilon^2.
$$
