Maximum of product of two functions

Question

Let's have two (real continuous differentiable) functions such that

1. $f(x)$ is bounded (from below and from above), positive ($f(x)>0$), and is strictly increasing ($f'(x)>0$, $\forall x$).
2. $g(x)$ is bounded (from below and from above) and has exactly one maximum ($g'(x_0) = 0$ ; $g(x)<g(x_0), \forall x \ne x_0$)

It then follows that function $h(x)=f(x)g(x)$ is also bounded. However, additional conditions on either $f(x)$ or $g(x)$ must apply in order for the function $h(x)$ to have again only one maximum.

The question is. What are the additional sufficient conditions on either $f(x)$ or $g(x)$ so that the product $h(x)=f(x)g(x)$ would have again only one maximum?

(Thanks to gammatester for pointing this out).

Answer 1

Dear @pisoir i think you need one easy condition or? read it.
$$h(x)=f(x)g(x)\\ h'(x)=f'(x)g(x)+f(x)g'(x)$$ The function $h(x)$ has a critical point if $h'(x)=0$, so $$h'(x)=f'(x)g(x)+f(x)g'(x)\\ =f'(x)\big\{ g(x)+\frac{f(x)}{f'(x)}*g'(x)\big\}$$ and we note that $f'(x)\ne0$ and that the fraction $\frac{f(x)}{f'(x)}$ is positive. The second derivative is $$h''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x) \space $$ we have the following cases(i will get the local maximum and discuss conditions to be a unique and hence global maximum):
(1) $x < x_0:$ in this interval both $f$ and $g$ are increasing and $f$ is strictly increase so there is no local maximum of $h$ in this inteval.
(2) $x = x_0$: $g'(x_0)=0$ and ang $g''(x_0)$ is negative, so comparing both $h'(x),h''(x)$, the only case that $h$ has a maximum is that $g(x_0)=0$. notice that in this case it will be a global maximum of $h(x)$ since $g(x)$ will be negative elsewhere
(3) $x \gt x_0$: in this interval $g'(x)$ is negative and the local maximum holds only if $$g(x)+\frac{f(x)}{f'(x)}*g'(x)=0$$ from this equation we gett that $g$ is positive and hence $h''$ is negative if both $f''$ and $g''$ are negative. this case needs a deep discussion and is rare for general functions to hold but you have a way to you complete.
A good graphical example for for two functions that have maximum product 10000 at infinity

Answer 2

When you are saying $g$ has just one maximum, I think it means that $g$ has no other point of maximum/minimum.

So the function increases to $g(x_0)$ and then decreases.

For $f$ it just goes on increasing.

$$ f(x)g(x) < f(x_0)g(x_0) \; \forall \; x \neq x_0 $$

Also check for $(f(x)g(x))'$ to be $0$ at $x = x_0$ we also need $g(x_0) = 0$.

So the conditions would be :

1) $ f(x)g(x) < f(x_0)g(x_0) \; \forall \; x \neq x_0 $ and

2) $g(x_0) = 0$.

which can be summarized in 2) only, i.e. $g(x_0) = 0$.

But we do not yet guaranteed that maxima is just one in this case.

If we need that there should be no other minima as well :

then $$ (f(x)g(x))' = 0 \; iff \; x = x_0 .$$

We analyse it over the 3 regions :

i) $x < x_0$ where $f'(x) > 0$ and $f(x) > 0$, $g(x) < 0$ and $g'(x) > 0$ ;

ii) $x = x_0$ and

iii) $x > x_0$ where $f'(x) > 0$ and $f(x) > 0$, $g(x) < 0$ and $g'(x) < 0$ ;

and ensure that in i) and ii) the condition does not hold.

$$ (f(x)g(x))' = f'(x)g(x) + g'(x)f(x) $$ is negative in region iii) so we do not have a min or maxima.

but in region i) the first term is negative while the last term is positive, so we have as all the quantities are positive,

$$f'(x)/g'(x) \neq - f(x)/g(x) \; for \; x < x_0.$$ for making the expression non-zero.

So the final conditions are :

1) $g(x_0) = 0$ and

2) $f'(x)/g'(x) \neq - f(x)/g(x) \; for \; x < x_0$.