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

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)

• Any function $f :\mathbb{R}_{+}\to\mathbb{R}_{+}$ is necessarily positive – Henry Apr 13 at 14:43
• Hint: A function with increasing derivative is convex. – Giorgos Giapitzakis Apr 13 at 14:45
• It slightly depends on your definition of derivative. $f(x) =x$ for $0 < x < 1$; $f(1)=3$; $f(x)=x+10$ for $x>1$ is a monotone increasing function but not obviously the derivative of its integral – Henry Apr 13 at 14:47
• you don't need positive. Check - $x^2$ has negative derivative for $x<0$, but you may want to add continuous to the requirement? – Rahul Madhavan Apr 13 at 14:48

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 $$\frac{F(y)-F(x)}{y-x} = \frac{1}{y-x} \int_x^y f(t) \, dt = \int_0^1 f(x + s(y-x)) \, ds \\ \le f(y) \le \int_0^1 f(y + s(z-y)) \, ds = \frac{F(z)-F(y)}{z-y} \, .$$
• If $f$ is continuous I understand that $F$ is convex, since $F'=f$ and $f$ is increasing. But what is the argument in the case $F'=f$ a.e.? – roi_saumon Apr 13 at 20:04
• Thank you. Is there a typo $f'\to f$? – roi_saumon Apr 13 at 23:22
• @roi_saumon: Yes, it should be $f$ (or $F'$). Hopefully correct now. – Martin R Apr 14 at 4:52