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:

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 $\map f x < \infty$.