Product of two monotonic real functions

Question

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 $f g$?

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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.

Answer 1

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.

Answer 2

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.