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:

\(\ds \map {V_{-g} } {\closedint a b}\)

\(=\)

\(\ds \size {-1} \map {V_g} {\closedint a b}\)

\(\ds \)

\(=\)

\(\ds \map {V_g} {\closedint a b}\)

so:

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

$\blacksquare$

Sources

• 1973: Tom M. Apostol: Mathematical Analysis (2nd ed.) ... (previous) ... (next): $\S 6.4$: Total Variation: Theorem $6.9$