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.
Best Answer
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.