Multiple of Absolutely Continuous Function is Absolutely Continuous

Theorem

Let $k$ be a real number.

Let $I \subseteq \R$ be a real interval.

Let $f : I \to \R$ be an absolutely continuous real function.

Then $k f$ is absolutely continuous.

Proof

Note that if $k = 0$, then $k f$ is constant.

Hence, by Constant Real Function is Absolutely Continuous:

$k f$ is absolutely continuous if $k = 0$.

Take now $k \ne 0$.

Let $\epsilon$ be a positive real number.

Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for every set of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:

$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$

we have:

$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {\size k}$

Then:

\(\ds \sum_{i \mathop = 1}^n \size {\map {\paren {k f} } {b_i} - \map {\paren {k f} } {a_i} }\)

\(=\)

\(\ds \size k \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }\)

\(\ds \)

\(<\)

\(\ds \size k \times \frac \epsilon {\size k}\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

whenever:

$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$

Since $\epsilon$ was arbitrary:

$k f$ is absolutely continuous if $k \ne 0$.

Therefore:

$k f$ is absolutely continuous for all $k \in \R$.

$\blacksquare$