Does strict convexity plus asymptotic affinity imply bounded mean?

Question

I am not sure if this is exactly research-level, but I am struggling to find a proof for the following claim:

Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function.

Let $\lambda_n \in [0,1],a_n\le c_0<b_n \in [0,\infty)$ satisfy $$ \lambda_n a_n +(1-\lambda_n)b_n=c_n $$ and suppose that $c_n \to c>c_0$.

Set $D_n=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(c_n\big) $, and suppose that $\lim_{n \to \infty}D_n=0$

Question: Must $b_n$ be bounded?

---

I have a quite simple proof (which I present below) for the special case where $a_n=a,c_n=c$ are constant sequences, but I am having trouble generalizing it.

The proof for the simplified case:

We have $ \lambda_n a +(1-\lambda_n)b_n=c$.

Given $x \ge r$, let $\lambda(x) \in [0,1]$ be the unique number satisfying $$ \lambda(x) a +(1-\lambda(x))x=c. $$ We have $\lambda(b_n)=\lambda_n$. Define $$g(x) = \lambda(x) F(a) + (1-\lambda(x))F(x).$$

The strict convexity of $F$ implies that $g$ is a strictly increasing function of $x$.

The assumption $D_n \to 0$ is equivalent to $g(b_n) \to F(c)$. Since $g(b_n) \ge F(c)$ (by convexity) and $g$ is strictly increasing, we conclude that $b_n$ must be bounded.

Answer 1

Yes, $b_n$ must be bounded. Assume the contrary. Passing to a subsequence we may suppose that $a_n\to a$, $b_n\to \infty$. We have $$\lambda_n=\frac{b_n-c_n}{b_n-a_n}\to 1;\, 1-\lambda_n=\frac{c_n-a_n}{b_n-a_n}\sim (c-a)b_n^{-1},$$ and using $F(b_n)\geqslant F(c_n)+(b_n-c_n)F'(c_n)$ we get $$ D_n+F(c_n)=\lambda_n F(a_n)+(1-\lambda_n)F(b_n)\geqslant \lambda_n F(a_n)+(1-\lambda_n)F(c_n)+(1-\lambda_n)(b_n-c_n)F'(c_n)\\ \to F(a)+(c-a)F'(c)>F(c), $$ thus $\liminf D_n>0$.