Product 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.

Then the pointwise product $f \cdot g$ is of bounded variation with:


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

where:


 * $\map {V_{f \cdot g} } {\closedint a b}$ denotes the total variation of $f \cdot g$ on $\closedint a b$
 * $A, B$ are non-negative real numbers.

Proof
For each finite subdivision $P$ of $\closedint a b$, write:


 * $P = \set {x_0, x_1, \ldots, x_n }$

with:


 * $a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$

By Function of Bounded Variation is Bounded:


 * $f$ and $g$ are bounded.

So, there exists $A, B \in \R$ such that:


 * $\size {\map f x} \le B$
 * $\size {\map g x} \le A$

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

Then:

Since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that:


 * $\map {V_f} {P ; \closedint a b} \le M$
 * $\map {V_g} {P ; \closedint a b} \le K$

for all finite subdivisions $P$.

We therefore have:


 * $\map {V_{f \cdot g} } {P ; \closedint a b} \le A M + B K$

so $f \cdot g$ is of bounded variation.

Further, we have: