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:

\(\ds \map {\paren {f \cdot g}^+} x\)

\(=\)

\(\ds \map f x \map g x\)

Definition of Positive Part

\(\ds \)

\(=\)

\(\ds \paren {-\map {f^-} x} \paren {-\map {g^-} x}\)

\(\ds \)

\(=\)

\(\ds \map {f^-} x \map {g^-} x\)

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:

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

$\blacksquare$