Positive 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 positive parts of $\alpha f$ and $f$ respectively
- $f^-$ is the negative 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:
- $\max \set {\alpha \map f x, 0} = \alpha \map f x$
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 $\alpha \map f x < 0$.
So:
- $\max \set {\map f x, 0} = 0$
and:
- $\max \set {\alpha \map f x, 0} = 0$
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 $\alpha \ge 0$.
Now take $\alpha < 0$.
Suppose that $\map f x \ge 0$.
Then $\alpha \map f x \le 0$.
So:
- $\max \set {\alpha \map f x, 0} = 0$
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 $\alpha \map f x > 0$.
So:
- $\max \set {\alpha \map f x, 0} = \alpha \map f x$
and:
- $-\min \set {\map f x, 0} = -\map f x$
So:
- $\map {\paren {\alpha f}^+} x = \alpha \map f x$
and:
- $\map {f^-} x = -\map f x$
So:
- $\map {\paren {\alpha f}^+} x = -\alpha \map {f^-} x$
if $\map f x < 0$.
So:
- $\paren {\alpha f}^+ = -\alpha f^-$
if $\alpha < 0$.
$\blacksquare$