Is my following reasoning right or wrong?

Question

I have three function $f_1(x)$, $f_2(x)$ and $f_3(x)$ (All are positive functions). The product of $f_1(x)$ and $f_2(x)$ is concave over $0<x<1$ while $f_3(x)$ is concave over $x_1<x<x_2$ where $0<x_1<1$ and $0<x_2<1$ and $x_2>x_1$. Can I say that the product of three functions $f(x)=f_1(x)f_2(x)f_3(x)$ is concave in $x_1<x<x_2$ (I am only interesting in maximizing $f(x)$) Can this function $f(x)$ has zero derivative in isolated locations in $x_1<x<x_2$? In my view the set over which the function has zero derivative should be convex since the product of positive functions has one maxima.

Answer 1

The product need not be concave. Let $f_1=1$,$f_2(x)=1-x$ and $f_3(x)=1-x^{2}$. Then $f_1 f_2 f_3$ is concave on $(0,1/3)$ and convex on $(1/3,1)$.

Answer 2

The product is not necessarily concave, and can have an isolated local extremum.

Take for example $x_1 = 0, \,x_2 = 1, \,f_1(x) = 1+x−x^2 ,f_2(x) = f_3(x) = 1−(2x−1)^2\,$, then:

• each of $f_1\,$, $f_2\,$, $f_3\,$ is positive and concave on $(0,1)\,$

• $f_1 \cdot f_2$ is also concave on $(0,1)$

• $f = f_1 \cdot f_2 \cdot f_3$ is not concave on $(0,1)\,$, in fact it changes concavity twice

• $f'$ has an isolated $0$ on $(0,1)$ which corresponds to a local maximum of $f$