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 $a_1<a_2<\ldots<a_n$ 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:
According to the above definition, 2-order convex is identical to convex.
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 $n=2$).
When $n=3$ this question is asking about the implication Q1 $\Longrightarrow$ Q2 in the question Geometric characterization of functions with positive third derivative.
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
