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?
> 
> 
> 




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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.



Does strict convexity plus asymptotic affinity imply bounded mean?
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 $\liminf D\_n>0$.


