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:
\(\ds \map f x \map g x\) | \(=\) | \(\ds \paren {\map {f^+} x - \map {f^-} x} \paren {\map {g^+} x - \map {g^-} x}\) | Difference of Positive and Negative Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f^+} x \map {g^+} x - \map {f^-} x \map {g^+} x - \map {f^+} x \map {g^-} x + \map {f^-} x \map {g^-} x\) |
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.
$\blacksquare$