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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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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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