Bound for Negative Part of Pointwise Sum of Functions
Theorem
Let $X$ be a set.
Let $f, g : X \to \overline \R$ be extended real-valued function.
Suppose that the pointwise sum $f + g$ is well-defined, that is:
- there exists no $x \in X$ such that $\set {\map f x, \map g x} = \set {\infty, -\infty}$.
Then:
- $\paren {f + g}^- \le f^- + g^-$
where $\paren {f + g}^-$, $f^-$ and $g^-$ denote the negative parts of $f + g$, $f$ and $g$ respectively.
Proof
Let $x \in X$.
From the definition of the negative part, we have:
- $\map {f^-} x = -\min \set {\map f x, 0}$
and:
- $\map {g^-} x = -\min \set {\map g x, 0}$
Suppose first that $\map f x$ and $\map g x$ are finite.
From Minimum Function in terms of Absolute Value, we then have:
- $\ds \map {f^-} x = \frac {\size {\map f x} - \map f x} 2$
and:
- $\ds \map {g^-} x = \frac {\size {\map g x} - \map g x} 2$
We then have:
| \(\ds \map {\paren {f + g}^-} x\) | \(=\) | \(\ds -\min \set {\map f x + \map g x, 0}\) | Definition of Negative Part | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\size {\map f x + \map g x} - \paren {\map f x + \map g x} } 2\) | Minimum Function in terms of Absolute Value | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \frac {\size {\map f x} + \size {\map g x} - \paren {\map f x + \map g x} } 2\) | Triangle Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\size {\map f x} - \map f x} 2 + \frac {\size {\map g x} - \map g x} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {f^-} x + \map {g^-} x\) |
Now suppose that $\map f x = +\infty$.
Then $\map g x > -\infty$.
We then have:
- $\map f x + \map g x = +\infty$
So:
- $\map {\paren {f + g}^-} x = 0$
and:
- $\map {f^-} x = 0$
Since:
- $\map {g^-} x \ge 0$
we have that:
- $\map {\paren {f + g}^-} x \le \map {f^-} x + \map {g^-} x$
Swapping $f$ for $g$, we also get the result if $\map g x = +\infty$ and $\map f x > -\infty$.
Now suppose that $\map f x = -\infty$
Then $\map g x < \infty$.
Then:
- $\map f x + \map g x = -\infty$
So:
- $\map {\paren {f + g}^-} x = \infty$
and:
- $\map {f^-} x = \infty$
So the inequality:
- $\map {\paren {f + g}^-} x \le \map {f^-} x + \map {g^-} x$
holds trivially.
Swapping $f$ for $g$, we also get the result if $\map g x = -\infty$ and $\map f x < \infty$.
$\blacksquare$