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So, non-vanishing of some derivatives does not ensure convexity. …

.$$ 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. …

The motivation is that I am applying Jensen inequality with $F$, and an affine part (in contrast to strict convexity) gives some flexiblity. …

The proof of the latter fact is not hard, but I couldn't find a source in the literature that presents this "localized" form of Jensen inequality, under the sole assumption of "convexity at a point". … Comment: Convexity at $c$ does not imply that the one-sided derivatives exist, so the standard proof for the existence of a supporting line (subgradient) does not apply here. …

My intuition is that even if $F$ becomes "less convex" (closer to being affine) when $x \to \infty$, then we cannot put to much weight on the $a_n$-since otherwise we get hit by the "convexity gap" between …

Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(1)=0$, and suppose that $|f|$ grows with the distance from $1$: $|f(x)|$ is strictly increasing when $x \ge 1$, and strictly dec …