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Here's a homework question I'm struggling with:

Prove/disprove the next statement:

Let $f,g$ two convex functions, then $h(x)=f(x) \cdot g(x)$ is also convex

So, we know that $h'(x)=f'(x) \cdot g(x) + f(x) \cdot g'(x)$. We also know that $f'(x),g'(x)$ are monotonically increasing because they are convex. If I can show that $h'(x)$ is also monotonically increasin I'm done, but I'm not sure how to do it. Any hints?

Thanks!

Community
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yotamoo
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  • Ok ... $x \mapsto (x-a)^2$ is convex for any $a$, right :) – martini May 21 '12 at 08:28
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    It would be nice to find some sufficient conditions under this statement is true. For example, maybe it's true when $f,g$ are non negative functions. – Joe Mar 27 '17 at 19:00
  • Necessary and sufficient conditions are given here: https://www.ima.umn.edu/sites/default/files/2204.pdf (despite the title it also covers the case of convex functions) – stewori Apr 18 '20 at 21:15

3 Answers3

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Hint: You can write the function defined by $x\mapsto-x^2$ as the product of two very simple linear and hence convex functions.

Michael Greinecker
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    So the statement is not even true. I assumed it is. Thanks! – yotamoo May 21 '12 at 08:37
  • I am sorry but I am new to convex functions. Is the statement "the product of convex functions is convex" true or false ? Could you please clarify and give more details. For example if $f(x)=-x^2$ and $g(x)=-x^2$ then $fg$ is a convex function whereas $f$ and $g$ are not. – roni Aug 25 '14 at 07:41
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    @roni It is false. Take the two functions $x\mapsto x$ and $x\mapsto -x$. – Michael Greinecker Aug 25 '14 at 16:40
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The functions $f(x)=1-x$ and $g(x)=1+x$ are convex. However, their product $(f*g)(x)=1-x^2$ is not. So it hoes not hold.

Censi LI
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Giannis
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5

False

Proof: Let $h(x) = f(x)\cdot g(x)$, or in short: $h = fg$, then: $h' = gf' + fg'$ and: $h'' = gf'' + 2f'g'+ fg''$

A necessary and sufficient condition for convexity is : $~h''= gf'' + 2f'g'+ fg'' \ge 0$ $\quad (1)$

A quick test is to check if $f$ and $g$ are both "$\ge 0$", and $f'$ and $g'$ have the same sign ("$\,\ge 0\,$": convex increasing; "$\,\le0\,$": convex decreasing). Otherwise, check condition $(1)$.

Note that if $f = g$, then $h= f^2$ and condition $(1)$ becomes:

$$h'' = 2ff'' + 2(f')^2$$

Since $f$ and $g$ are convex, $f''\ge 0$ and $g'' \ge 0$, then the only condition to check is $f \ge 0$.

them
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Hass
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  • Onother user asked a sufficient condition for the product of convex functions to be convex. Your calculation of second derivatives shows that if both functions are increasing, positive and convex, then the product function is increasing positive and convex. Maybe it is possible two show this result without mentioning to derivatives. – R. W. Prado Nov 12 '24 at 22:09