This is a follow-up of this question.
Let $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$.
Suppose also that $F|_{(1-\epsilon,1+\epsilon)}$ is convex and smooth for some $\epsilon>0$.
Question: Is it true that $F$ is convex at at every point $y \in (1-\epsilon,1)$?
By convexity at a point $y$, I mean that that for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =y$, we have $$ F(y)=F\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha F(x_1) + (1-\alpha)F(x_2). \tag{1} $$ Equivalently, $F$ is above its tangent at $(y,F(y))$: $$ F(x) \ge F(y)+F'(y) (x-y) \tag{2} $$ for every $x \in (0,\infty)$.
I know that there exist a $\delta>0$ such that $F$ is convex at every point $y \in (1-\delta,1)$. The question is whether we can take $\delta=\epsilon$.
Counterexamples can be given. Suppose $F$ can be continuously extended to $0$ and the property holds. Let $x_n\to0$ and let $t_n= \frac{y-x_n}{1-x_n}$ so that $(1-t_n)x_n + t_n1=y$. As $n\to\infty$ you get that $$(1-t_n)F(x_n)+t_nF(1)=(1-t_n)F(x_n)\to (1-y) F(0)$$ by assumption each term is $≥F(y)$, hence the limit also is larger than $F(y)$ and you get that $$F(0)≥\frac{F(y)}{1-y}$$ must necessarily hold. But by choosing $F(x)=(x-1)^2$ on $[\frac12,\infty)$ and then extending smoothly and monotonously on $[0,\frac12]$ so that $F(0)=\frac38$ you find that all conditions are fufilled but the necessary condition derived above fails for $y=\frac12$. This basic idea can be applied also if you are interested in $F$ being divergent at $0$.