# Product of two monotonic real functions

I'm confused about the possibility to say that the product of two monotonic real functions $f,g:\mathbb{R} \to \mathbb{R}$ is monotonic.

I found the following proposition:

A. If $g$ and $f$ are two increasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is increasing.

B. If $g$ and $f$ are two increasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is decreasing.

Does something similar hold also for decreasing functions? That is: is the following sentence correct?

A'. If $g$ and $f$ are two decreasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is decreasing.

B'. If $g$ and $f$ are two decreasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is increasing.

Moreover, can something be said about the product of a increasing and a decreasing functions?

For example if $f$ is increasing and $g$ is decreasing under what conditions can I say something about the monotony of the product $f g$?

Besides these two practical questions I would like to ask some suggestions on how to prove statement A.

I tried in the following way

$Hp:$

$x_1 >x_2 \implies f(x_1)>f(x_2)>0 \,\,\, \forall x_1,x_2$

$x_1 >x_2 \implies g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2$

$Th:$

$x_1 >x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$

$Proof:$

$f(x_1)>f(x_2)>0 \,\,\, , g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$

Which seems obvious if one thinks about some numbers but I don't really know how I could prove the last implication in rigourous way. So any help in this proof is highly appreciated.

To prove statement A, let $x > y$. Then, as we know, $f(x) > f(y)>0$ and $g(x) > g(y)>0$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)>0)\\ g(x) > g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}

An analogous proof would follow for part B if $f$ and $g$ were increasing, with a caveat:let $x > y$. Then, as we know, $f(x) > f(y)$ and $g(x) > g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)<0)\\ g(x) > g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}

Now, let us see if the same logic could work with part A':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)>0)\\ g(x) < g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}

That's brilliant, so great intuition for anticipating part A'. Now we will check part B':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)<0)\\ g(x) < g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}

And therefore part B' is also done. Note the above logic carefully, I think all steps are equally important.

Use this logic, and see why in most cases, one increasing and one decreasing function doesn't tell you much about the product itself.

For A' and B', apply A and B. For A': If $f,g$ are decreasing and positive then $-f$ and $-g$ are increasing and negative so by B, the function $(-f)(-g)=fg$ is decreasing. Similarly, apply A to B'.

$e^x$ is increasing and $e^{-x}+1$ is decreasing, and their product $1+e^x$ is increasing.

$e^x+1$ is increasing and $e^{-x}$ is decreasing, and their product $1+e^{-x}$ is decreasing.

The product of a positive increasing and a positive decreasing function can also fail to be monotonic.