Product of increasing functions need not be increasing

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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 \R.

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



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.