Product of Positive and Negative Parts is Zero

Theorem

Let $X$ be a set.

Let $f : X \to \R$ be a real-valued function.

Let $f^+$ and $f^-$ be the positive part and negative part of $f$ respectively.

Then $f^+ f^- = 0$.

Proof

Let $x \in X$.

If $\map f x = 0$, then we have $\map {f^+} x = \map {f^-} x = 0$.

Hence $\map {f^+} x = \map {f^-} x = 0$ in this case.

If $\map f x > 0$, then $\map {f^+} x = \map f x$ and $\map {f^-} x = 0$.

So $\map {f^+} x \map {f^-} x = 0$ in this case also.

Finally, if $\map f x < 0$, then $\map {f^+} x = 0$ and $\map {f^-} x = -\map f x$.

So $\map {f^+} x \map {f^-} x = 0$ in this case.

We have covered all possibilities, so we have $\map {f^+} x \map {f^-} x = 0$ for each $x \in X$.

$\blacksquare$