every positive monotone increasing function necessarily the derivative of a convex function?

Question

I was trying to find out if every positive (do we need this?) monotone increasing function $f :\mathbb{R}_{+}\to\mathbb{R}_{+}$ was necessarily the derivative of a convex function. I have found this answer that seems related but I am not so sure if this is the good approach. Do I have to verify that for $n$ points $x_1,\ldots, x_n$ I have $\langle x_0-x_1,f(x_0)\rangle+\ldots+\langle x_{n-1}-x_n,f(x_{n-1})\rangle+\langle x_{n}-x_0,f(x_n)\rangle\ge 0$? (for every $f$ as above)

Answer 1

If $F: I \to \Bbb R$ is convex and differentiable on an interval $I \subset \Bbb R$ then $F'$ is increasing and continuous, see for example Continuity of derivative of convex function.

Conversely, if $f: I \to \Bbb R$ is increasing and continuous then $$ \tag{*} F(x) = \int_a^x f(t) \, dt $$ (for some $a \in I$) is convex with $F'=f$.

If $f$ is only increasing but not necessarily continuous then we can still define $F$ via $(*)$ because monotone functions are Lebesgue integrable. Then $F' = f$ a.e. in $I$ and $F$ is convex: For $x < y < z$ is $$ \frac{F(y)-F(x)}{y-x} = \frac{1}{y-x} \int_x^y f(t) \, dt = \int_0^1 f(x + s(y-x)) \, ds \\ \le f(y) \le \int_0^1 f(y + s(z-y)) \, ds = \frac{F(z)-F(y)}{z-y} \, . $$