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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).

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  • $\begingroup$ Let $f(x) = \tan^{-1} x$, $g(x) = -x^2$. These satisfy your criteria, but the product, being an odd function, does not. $\endgroup$ – heropup Jan 21 '14 at 8:35
  • $\begingroup$ @heropup I edited the question. $g(x)$ must be also bounded from below. $-x^2$ is not. $\endgroup$ – pisoir Jan 21 '14 at 8:39
  • $\begingroup$ if $g'(x)<g'(x_0)=0$ for each $x\neq x_0$ then it has no maximum at $x_0$ and is strictly decreasing. To get a maximum the sign of the derivative must switch. You need $g'(x)>0$ if $x<x_0$ and $g'(x)<0$ if $x>x_0$. $\endgroup$ – drhab Jan 21 '14 at 8:42
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    $\begingroup$ If the function is odd, then any local extrema are symmetric on the real line. This behavior would require at least two critical points, but it is easy to construct examples where we don't have that many, as I have shown. It's not intended to be a rigorous argument or strict criterion. $\endgroup$ – heropup Jan 21 '14 at 9:07
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    $\begingroup$ You must put even stronger restrictions on $f,g.\;$ With $$f(x)=\arctan(x)+2,\quad g(x)=\frac{1}{\arctan(x)^2-4}$$ all your requirements are met, but $$f(x)g(x)=\frac{1}{\arctan(x)-2}$$ has no maximum. $\endgroup$ – gammatester Jan 21 '14 at 9:24
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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 An example

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  • $\begingroup$ I was hoping for a more "substantial" condition which is easy to check directly. Like (just a pure example) that if $g(x)$ is Lipschitz continuous (or anything) then it holds. $\endgroup$ – pisoir Feb 3 '14 at 11:45
  • $\begingroup$ I was wondering. Could you maybe show, using your method, why for e.g. (see gammatester comment above) $f(x)=\arctan(x)+2$, $g(x)=\frac{1}{(\arctan(x))^2-4}$ there is no maximum for $h(x)=f(x)g(x)$? $\endgroup$ – pisoir Feb 3 '14 at 13:31
  • $\begingroup$ (First i didn't check the conditions of both functions, i assumed it holds). The first case to have a maximum is $g(x_0)=0$ is not valid since $g(0)=-1/4$ $\endgroup$ – Semsem Feb 3 '14 at 14:41
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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) $f'(x)/g'(x) \neq - f(x)/g(x) \; for \; x < x_0$.

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  • $\begingroup$ the first one is incorrect ingeneral $\endgroup$ – Semsem Feb 3 '14 at 11:49
  • $\begingroup$ @Semsem : Good now ? $\endgroup$ – DiffeoR Feb 3 '14 at 11:59
  • $\begingroup$ I think no @Diffeor $\endgroup$ – Semsem Feb 3 '14 at 12:24
  • $\begingroup$ @Semsem : Probably this is the best to offer if there are no serious mistakes. You have assumed double differentiability (an extra assumption) in your proof which I have tried to avoid here. $\endgroup$ – DiffeoR Feb 3 '14 at 12:36
  • $\begingroup$ @pisoir : Is it simple enough now ? $\endgroup$ – DiffeoR Feb 3 '14 at 12:40

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