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:


 * $\map {\paren {f \cdot g}^-} x = \begin{cases}\map {f^-} x \map {g^+} x & \map f x \le 0 \text { and } \map g x \ge 0 \\ \map {f^+} x \map {g^-} x & \map f x \ge 0 \text { and } \map g x \le 0 \\ 0 & \text {otherwise}\end{cases}$

for each $x \in X$, where:


 * $f \cdot g$ is the pointwise product of $f$ and $g$
 * $\paren {f \cdot g}^-$ denotes the negative part.

Proof
Let $x \in X$ be such that $\map f x \le 0$ and $\map g x \ge 0$.

Then, we have:


 * $\map f x \map g x \le 0$

So:


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

by the definition of the negative part.

Then, from the definition of the negative part, we have:


 * $\map {f^-} x = -\map f x$

From the definition of the positive part, we have:


 * $\map {g^+} x = \map g x$

Then, we have:

Now let $x \in X$ be such that $\map f x \ge 0$ and $\map g x \le 0$.

Swapping $f$ for $g$ in the previous computation gives:


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

Now let $x \in X$ be such that neither:


 * $\map f x \ge 0$ and $\map g x \le 0$

or:


 * $\map f x \le 0$ and $\map g x \ge 0$

hold.

Then:


 * $\map f x > 0$ and $\map g x > 0$

or:


 * $\map f x < 0$ and $\map g x < 0$.

Then:


 * $\map f x \map g x > 0$

So by the definition of the negative part, we have:


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