Negative Part of Pointwise Product of Functions

Theorem

Let $X$ be a set.

Let $f, g : X \to \overline \R$ be extended real-valued functions.

Then:

$\paren {f \cdot g}^- = f^- g^+ + f^+ g^-$

where:

$f \cdot g$ is the pointwise product of $f$ and $g$

$\paren {f \cdot g}^-$ denotes the negative part.

We have:

$\ds \map f x \map g x$

$=$

$\ds \paren {\map {f^+} x - \map {f^-} x} \paren {\map {g^+} x - \map {g^-} x}$

$\ds $

$=$

$\ds \map {f^+} x \map {g^+} x - \map {f^-} x \map {g^+} x - \map {f^+} x \map {g^-} x + \map {f^-} x \map {g^-} x$

Note that we have $\map f x \map g x \le 0$ if and only if $\map f x$ and $\map g x$ have the opposite sign.

That is, if and only if $\map f x \ge 0$ and $\map g x \le 0$ or $\map f x \le 0$ and $\map g x \ge 0$.

In the first case, we have $\map {g^+} x = 0$ and $\map {f^-} x = 0$, so:

$\map {f^+} x \map {g^+} x = \map {f^-} x \map {g^-} x = 0$

In the second case, we have $\map {f^+} x = 0$ and $\map {g^-} x = 0$, so:

$\map {f^+} x \map {g^+} x = \map {f^-} x \map {g^-} x = 0$

So, in either case we have:

$\map f x \map g x = -\paren {\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x}$

so since:

$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x \ge 0$

we obtain:

$\map {\paren {f \cdot g}^-} x = \map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x$

if $\map f x \map g x \le 0$.

If $\map f x \map g x \ge 0$, $\map f x$ and $\map g x$ have the same sign.

That is, either $\map f x \ge 0$ and $\map g x \ge 0$ or $\map f x \le 0$ and $\map g x \le 0$.

In the first case, we have $\map {f^-} x = 0$ and $\map {g^-} x = 0$, so:

$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x = \map {\paren {f \cdot g}^-} x = 0$

In the second case, we have $\map {f^+} x = 0$ and $\map {g^+} x = 0$, so:

$\map {f^-} x \map {g^+} x + \map {f^+} x \map {g^-} x = \map {\paren {f \cdot g}^-} x = 0$

and hence we are done.

$\blacksquare$