Positive 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:


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

for each $x \in X$, where:


 * $f \cdot g$ is the pointwise product of $f$ and $g$
 * $\paren {f \cdot g}^+$ denotes the positive part

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

Then:


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

From the definition of the positive part, we then have:


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

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

Then, we have:


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

and:


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

from the definition of the negative part.

Then, we have:


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

So again:

Now suppose that $x \in X$ is such that neither of:


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

or:


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

are true.

Then, we have either $\map f x > 0$ and $\map g x < 0$, or $\map f x < 0$ and $\map g x > 0$.

In either case:


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

So, from the definition of the positive part, we have:


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