2

It is generally known that multiplication of convex functions is not convex. Take for example:

$f_1(x) = 1-x$, $f_2(x) = 1+x$, Then $f_1(x)f_2(x) = 1-x^2$

Which is not convex

However, I've encountered situations where convexity appears to hold true in specific scenarios. Take for example:

$f(x) = a^xb^x$ where $a, b >=0$

Can this be generalized to a broader class of functions? Say for example strictly-convex functions/non-decreasing/non-increasing, a composition of requirements, or something more general. Much would be appreciated even if it is a link to a theorem for said specific case. Thanks!

rwalsh3750
  • 77
  • I also found this I don't think it is a dupe, as it simply ask to prove that the multiplication of two convex functions is not convex – Marco Feb 10 '24 at 18:41
  • 1
    Yes, there are multiple examples on this site asking about the product of two specific functions. Like you said, these and some others partly answer the question, but not generally. – rwalsh3750 Feb 10 '24 at 18:50

1 Answers1

2

Suppose $f,g$ are twice differentiable, convex, both not increasing or not decreasing and positive. Than $fg$ is convex

Indeed by the generalised Leibnitz rule we have $$ \frac{d^2}{dx^2}(fg)= f''g+ f g'' + 2f'g' \ge 0 $$ As $f'',g'' \ge 0$ by convexity, $f'g' \ge 0$ because the two functions have the same monotonicity and $f,g \ge0$ by hypothesis. (at least in one dimension the condition of being twice differentiable is not necessary. See exercise 3.32 (a) of convex optimization, S. Boyd, L. Vandenberghe )

Another sufficient (but not necessary) condition is that the two function are log convex as in that case you can write the product as $$ fg(x)=e^{\log(f(x))+\log(g(x))} $$ As the sum of two convex function is convex the product is log-convex, so it is convex (see e.g. this)

I don't think one can do much better than this.

Marco
  • 2,357