Product of increasing functions need not be increasing

Statement

It is possible to have two real-valued functions $f,g$, both defined on all real numbers, such that both $f$ and $g$ are increasing functions but the product $fg$ (defined as $x\mapsto f(x)g(x)$) is not an increasing function on all of $\mathbb {R}$.

We can also adapt this to functions defined on any open interval or closed interval instead of all real numbers.

Proof

Counterexample

Since we need to disprove a general statement, it is enough to exhibit one counterexample.

The example is with $f,g$ both the identity function:

$\!f(x):=x,g(x):=x$

These are both increasing on all real numbers.

The product is:

$\!(fg)(x)=f(x)g(x)=xx=x^{2}$

This is the square function which is decreasing on $(-\infty ,0)$ and increasing on $(0,\infty )$.

Generic idea behind counterexample

A product of negative increasing functions is decreasing.

Compatibility with product rule for differentiation

The product rule for differentiation states that:

$\!(fg)'(x)=f'(x)g(x)+f(x)g'(x)$

In particular, the sign of $(fg)'$ depends not merely on the signs of $f'$ and $g'$ but also on the signs of $f$ and $g$. In particular, if $g$ is negative, then multiplying by a positive $f'$ gives a negative product. This is why $(fg)'$ can be negative even though $f',g'$ are both positive.