# A higher-order analogy to that if $f'$ is increasing on $\Bbb R$ except for countably many points, then $f$ is convex

It is well known that if a continuous function $$f:\Bbb R \to \Bbb R$$ is differentiable at all but countably many points and the derivative it increasing on its domain, then $$f$$ is convex. Does its extension to "higher-order" convexity hold?

Consider $$n \in \Bbb N_{\geq 2}$$. If a function $$f$$ is $$n-2$$ times continuously differentiable and $$n-1$$ times differentiable at all but countably many points and $$f^{(n-1)}$$ is increasing on its domain then $$f$$ is $$n$$-order convex.

Definition: We say that a function $$f:\Bbb R \to \Bbb R$$ is $$n$$-order convex iff for all $$x\in \mathbb R$$ iff for any $$a_1 the $$n-1$$ degree polynomial $$p(x)$$ interpolating the points $$\big(a_1,f(a_1)\big),\ldots,\big(a_n,f(a_n)\big)$$ is such that the graph of its restriction $$p$$ to $$[a_{n-1},a_n]$$ is contained in the epigraph $$\mathop{epi}(f)$$.

Notes:

1. According to the above definition, 2-order convex is identical to convex.

2. I got this question when trying relax the assumption that $$f$$ is $$n$$ times differentiable in the result: Geometric characterization of the $n$-th derivative of $f$ being positive (convexity for $n=2$).

3. When $$n=3$$ this question is asking about the implication Q1 $$\Longrightarrow$$ Q2 in the question Geometric characterization of functions with positive third derivative.

4. The reverse implication is discussed in the post A higher-order analogy to that every convex function is differentiable at all but countably many points

• The statement in the first pargraph is false unless $f$ is assumed tio be continuous, Commented Jan 14, 2023 at 8:55
• @geetha290krm Thanks a lot for pointing it out! I also edited the statement of the claim. Commented Jan 14, 2023 at 9:19