Product of Absolutely Continuous Functions is Absolutely Continuous

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Theorem

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

Let $f, g : \closedint a b \to \R$ be absolutely continuous real functions.


Then $f \times g$ is absolutely continuous.


Proof

From Absolutely Continuous Real Function is Continuous:

$f$ and $g$ are continuous.

From Closed Real Interval is Compact in Metric Space:

$\closedint a b$ is compact.

Therefore, by Continuous Function on Compact Subspace of Euclidean Space is Bounded:

$f$ and $g$ are bounded.

That is, there exists $M_f, M_g \in \R_{> 0}$ such that:

$\size {\map f x} \le M_f$
$\size {\map g x} \le M_g$

for all $x \in \closedint a b$.

Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a set of disjoint closed real intervals.

Then:

\(\ds \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} }\) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} \map g {b_i} - \map f {a_i} \map g{a_i} + \map f {a_i} \map g {b_i} - \map f {a_i} \map g {b_i} }\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} \paren {\map f {b_i} - \map f {a_i} } + \map f {a_i} \paren {\map g {b_i} - \map g {a_i} } }\)
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} } \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map f {a_i} } \size {\map g {b_i} - \map g {a_i} }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds M_g \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + M_f \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }\) since $a_i, b_i \in \closedint a b$

Let $\epsilon$ be a positive real number.

Since $f$ is absolutely continuous, there exists $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:

$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$

we have:

$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$

Similarly, since $g$ is absolutely continuous, there exists $\delta_2 > 0$ such that whenever:

$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$

we have:

$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$

Let:

$\delta = \map \min {\delta_1, \delta_2}$

Then, whenever:

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

We have both:

\(\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }\) \(<\) \(\ds \frac \epsilon {2 M_g}\)
\(\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }\) \(<\) \(\ds \frac \epsilon {2 M_f}\)

and hence:

\(\ds \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} }\) \(<\) \(\ds M_g \times \frac \epsilon {2 M_g} + M_f \times \frac \epsilon {2 M_f}\)
\(\ds \) \(=\) \(\ds \epsilon\)

Since $\epsilon$ was arbitrary, we have:

$f \times g$ is absolutely continuous.

$\blacksquare$