Negative Part of Multiple of Function

Theorem
Let $X$ be a set.

Let $f : X \to \overline \R$ be an extended real-valued function.

Let $\alpha$ be a real number.

Then:


 * $\ds \paren {\alpha f}^- = \begin{cases}\alpha f^- & \alpha \ge 0 \\ -\alpha f^+ & \alpha < 0\end{cases}$

where:
 * $\paren {\alpha f}^-$ and $f^-$ are the negative parts of $\alpha f$ and $f$ respectively
 * $f^+$ is the positive part of $f$.

Proof
Let $x \in X$.

First take $\alpha \ge 0$.

Suppose that $\map f x \ge 0$.

Then $\alpha \map f x \ge 0$.

So:


 * $-\min \set {\alpha \map f x, 0} = 0$

and:


 * $-\min \set {\map f x, 0} = 0$

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


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

and:


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

So:


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

if $\map f x \ge 0$.

Now suppose that $\map f x < 0$.

Then $\alpha \map f x < 0$.

Then:


 * $-\min \set {\map f x, 0} = -\map f x$

and:


 * $-\min \set {\alpha \map f x, 0} = -\alpha \map f x$

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


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

and:


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

So:


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

if $\map f x < 0$.

So:


 * $\paren {\alpha f}^- = \alpha f^-$

if $\alpha \ge 0$.

Now take $\alpha < 0$.

Suppose that $\map f x \ge 0$.

Then $\alpha \map f x \le 0$.

Then:


 * $-\min \set {\alpha \map f x, 0} = -\alpha \map f x$

and:


 * $\max \set {\map f x, 0} = \map f x$

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


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

and:


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

So:


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

if $\map f x \ge 0$.

Now suppose that $\map f x < 0$.

Then $\alpha \map f x > 0$.

We then have:


 * $-\min \set {\alpha \map f x, 0} = 0$

and:


 * $\max \set {\map f x, 0} = 0$

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


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

and:


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

So:


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

if $\map f x < 0$.

So:


 * $\paren {\alpha f}^- = -\alpha f^+$

if $\alpha \ge 0$.