Proving that multiplication of convex function is convex

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

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.

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 $f \ge 0$.