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Problem: Let $f,g: [0, + \infty[ \to \mathbb{R}$ be two convex functions of Class $C^2$. Assume that $$ f(0) \geq 0, g(0) \geq 0 \text{ and } f'(0) \geq 0, g'(0) \geq 0 \tag{!}$$ Show that $fg$ is convex

My approach: I don't really understand what to do with the given information tagged as (!). I do however know the following:

I focus on $f$, the same will apply for $g$. Since $f$ is a convex function of Class $C^2$ and we have $f'(0) \geq 0$ there must exists some positive $\delta_1>0$ for which we have $$f': [0, \delta_1] \to \mathbb{R} \text{ is increasing } \implies f'' \geq 0, \ \forall x \in [0, \delta_1] $$

Analogous for $g$ with some different $\delta_2 >0$. Define $\delta =\min(\delta_1,\delta_2)$ such that both $f$ and $g$ are on the Intervall $I=[0,\delta]$ simultaneously convex meaning: $$f'' \geq 0 \text{ and } g''>0,  \forall x \in [0,\delta] $$


The above was merely some pre work I did to understand the problem better, not sure if it was necessarily.

For the 2nd part I define $h: I=[0, \delta] \to \mathbb{R}, \ h \mapsto fg \in C^2$ because it is a composition of $C^2$ functions. Now I differentiate $h$ twice: $$h'=f'g+g'f \implies h''=f''g+g'f'+g''f+f'g' \\ \implies h''=f''g+2f'g'+g''f \geq 0, \forall x \in I $$

Which would show that $h$ is convex. It looks correct to me but I'd like to hear some opinions if my approach was correct or if I have some flawed thinking at some points. Please comment/answer to highlight my mistakes or suggest improvisations.

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1 Answer 1

up vote 1 down vote accepted

You have $h'' = f''g + 2f'g' + g''f$, but you haven't explained why $f''g + 2f'g' + g''f \geq 0$. You can assert this if you know that $f''$, $g''$, $f'$, $g'$, $f$, and $g$ are all positive on $[0,\infty)$. You know that $f''$ and $g''$ are positive by convexity. The information in (!) is supposed to help you show that $f'$, $g'$, $f$, and $g$ are all positive. Then you will be done.

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Thanks for your answer @AlexZorn, I was trying to empathize that the 2nd derivatives of $f,g$ are strictly positive on the interval I have defined. But that seems to have been a redundant step because, as you said, they are convex. I do however believe that my defined pre work shows that $f',g',f,g$ are positive on the interval $I=[0, \delta]$ –  Spaced Apr 30 at 14:43

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