If two functions are increasing is their product increasing?

Question

The problem is:

If $f$ and $g$ are increasing, then is $f \cdot g$ also increasing?

First, I started out by working some basic definitions and assumptions:

• Assume $a < b$
• Increasing means $f(a) < f(b)$

Then I tried using case analysis:

Case 1: Assume $f(a), f(b), g(a), g(b)$ are all positive. Then:

$$f(a)g(a) < f(b)g(b)$$

which means $f \cdot g$ is increasing.

Case 2: Assume $f(a), f(b), g(a), g(b)$ are all negative. Then:

$$f(a)g(a) > f(b)g(b)$$

which means $f \cdot g$ is not increasing.

Case 3: Assume $f(a), f(b)$ are positive and $g(a), g(b)$ are negative. Then:

$$f(a)g(a) > f(b)g(b)$$

Means $f \cdot g$ is not increasing.

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A few problems that I don't like with my approach:

• The case analysis is quite cumbersome, and I still miss cases where the functions are equal to $0$.

• There is no real proof here, I mostly tried with numerical examples and general logic. Is there an algebraic way to start with the inequality $f(a) < f(b)$ and then modify it so that we end up with $f(a)g(a) < f(b)g(b)$? I couldn't get anywhere with that approach though.

Answer 1

The statement is false and a single example will prove it. Take $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.

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Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $\mathbb R$).

Answer 2

Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand, $$ f(x)g(x)=x^2-1 $$ is decreasing over $(-\infty,0]$ and increasing over $[0,\infty)$.

What's the problem? In the interval $(-\infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so $$ f(a)g(a)>f(b)g(b) $$ Similarly, the product is certainly increasing where both functions are positive.

In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.

You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.