Multiple of Function of Bounded Variation is of Bounded Variation

Theorem

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

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

Let the total variation of $f$ on $\closedint a b$ be $\map {V_f} {\closedint a b}$.

Then $k f$ is of bounded variation with:

$\map {V_{k f} } {\closedint a b} = \size k \map {V_f} {\closedint a b}$

where $\map {V_{k f} } {\closedint a b}$ is the total variation of $k f$ on $\closedint a b$.

Proof

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

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

with:

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

Then:

\(\ds \map {V_{k f} } {P ; \closedint a b}\)

\(=\)

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

using the notation from the definition of bounded variation

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \size k \map {V_f} {P ; \closedint a b}\)

Since $f$ is of bounded variation, there exists $M \in \R$ such that:

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

for all finite subdivisions $P$.

So:

$\map {V_{k f} } {P ; \closedint a b} \le \size k M$

So $k f$ is of bounded variation.

We then have:

\(\ds \map {V_{k f} } {\closedint a b}\)

\(=\)

\(\ds \sup_P \paren {\map {V_{k f} } {P ; \closedint a b} }\)

Definition of Total Variation of Real Function

\(\ds \)

\(=\)

\(\ds \sup_P \paren {\size k \map {V_f} {P ; \closedint a b} }\)

\(\ds \)

\(=\)

\(\ds \size k \sup_P \paren {\map {V_f} {P ; \closedint a b} }\)

Multiple of Supremum

\(\ds \)

\(=\)

\(\ds \size k \map {V_f} {\closedint a b}\)

$\blacksquare$