I'm aware that if $h(x)$ and $f(x)$ are convex functions, $g(x) = h(x)f(x)$ may not necessarily be convex.
I'm curious whether $g(x)$ is convex if both $h(x)$ and $f(x)$ are also monotonically decreasing along being convex, e.g. $f = e^{-x}$. I couldn't come up with a counter-example? Maybe there is such a property?
If $h$ and $f$ are both positive functions and if both are increasing or decreasing then their multiplication is also increasing or decreasing.
On $f,h\in\mathbb{R}$ this is no more true. Take some continuous versions of these $f:=(2,1,0.5)$ and $h:=(-0.5,-0.8,-1)$ on some subset of $\mathbb{R}$, then $fh=(-1,-0.8,-0.5)$ which is increasing. The second order differences of both are positive, so the corresponding continuous versions will be convex.
In the same spirit let $f(x)=e^{-2x}$ and take $h(x)=-10x$ on $\mathbb{R}^+$ and $h(x)$ is something else (convex and decreasing) on $\mathbb{R}^-$ both are convex and decreasing but $f(x)g(x)=-10xe^{-2x}$ is not convex.
