Difference of Functions of Bounded Variation is of Bounded Variation

Theorem
Let $a, b$ be real numbers with $a < b$.

Let $f, g : \closedint a b \to \R$ be functions of bounded variation.

Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$.

Then $f - g$ is of bounded variation with:


 * $\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$

where $V_{f - g}$ denotes the total variation of $f - g$ on $\closedint a b$.

Proof
By Multiple of Function of Bounded Variation is of Bounded Variation, we have that:


 * $-g$ is of bounded variation.

So, by Sum of Functions of Bounded Variation is of Bounded Variation, we have that:


 * $f + \paren {-g} = f - g$ is of bounded variation

with:


 * $\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_{-g} } {\closedint a b}$

where $V_{-g}$ is the total variation of $-g$ on $\closedint a b$.

We have, by Multiple of Function of Bounded Variation is of Bounded Variation:

so:


 * $\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$