# A sufficient condition for a unique maximum of the product of two concave functions

Given two concave functions $f(x)$ and $g(x)$, what conditions in terms of these functions can ensure that

$h(x)=f(x)g(x)$

have a unique maximizer on an interval $[a,b]$ for $a<b$?

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Please, do not use $*$ to denote multiplication between two real numbers. – Siminore Oct 25 '12 at 9:23
Now it looks like a good question. Just a nit, but how about replacing the word "maximizer" by the more standard term "maximum"? That is, you want $f(x)g(x)$ to have exactly one absolute maximum on $[a,b]$. Also, do you also want there not to be any local maxima other thant the absolute one? – coffeemath Oct 25 '12 at 11:53

Here's one way: Let $h(x)=f(x)g(x)$ and make the following assumptions about the concave functions $f,g$ on $[a,b]$:

[1] $f,g$ are nonnegative on $[a,b]$

[2] $f(a)g(a)=f(b)g(b)$

[3] $f'(x)g'(x)<0$ on $(a,b)$.

Then since $h(a)=h(b)$, Rolle's theorem says there is $c$ in $[a,b]$ for which $h'(c)=0$, and by the assumptions $f''(x)<0$ and $g''(x)<0$ [concavity of $f,g$], and using $f'(x)g'(x)<0$, a look at the second derivative

$h''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$

shows it to be negative on $(a,b)$, so that we have a local max at $c$. And since $h''(x)<0$ everywhere in $(a,b)$ we see that $h$ is concave down, and so can have only the one local max at $c$, which is therefore the unique local max and the unique global max.

An example of such $f,g,I$ is $f(x)=\sqrt{x},g(x)=\sqrt{1-x}$ on the interval $I=[0,1]$. Then the conditons 1,2,3 hold, and there is a unique maximum at $1/2$.

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