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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 $\delta=\epsilon$.

Asaf Shachar
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1 Answers1

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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 $0$.

s.harp
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  • Thanks, this is a very cool example. Do you really think that an analogous idea should work for $F$ which is divergent at $0$? – Asaf Shachar Jul 13 '20 at 11:38
  • Yes, you can get an analogous inequality for the value of something like $F(1/2)$, then let $F$ be quadratic on $[3/4,\infty)$ and modify $F$ suitably from $3/4$ to $1/2$ so that the inequality is violated. To the left of $1/2$ let $F$ diverge. Using $0$ and $1$ just makes the inequality look pleasing. – s.harp Jul 13 '20 at 11:42
  • Alternatively note that if you rescale the argument of a convex thing it remains convex, and if you translate the argument it will also remain convex. So extend the $F$ above to $-1$ so that it diverges at $-1$ and then let $\tilde F(x) := F( \frac12 x + \frac12)$, which now diverges at $0$, satisfies the requirements but fails to be convex at $y=3/4$. – s.harp Jul 13 '20 at 12:24