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.

Proof
We have:

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

That is, $\map f x \ge 0$ 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.