# Monotone Function is of Bounded Variation

## Theorem

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

Let $f : \closedint a b \to \R$ be a monotone function.

Then $f$ is of bounded variation.

## Proof

We use the notation from the definition of bounded variation.

Let $P = \set {x_0, x_1, \ldots, x_n}$ be a finite subdivision of $\closedint a b$.

As $f$ is monotone, it is either increasing or decreasing.

First consider the case of $f$ increasing, then:

$\map f {x_i} \ge \map f {x_{i - 1} }$

for all $i \in \N$ with $i \le n$.

By the definition of the absolute value, we have:

$\size {\map f {x_i} - \map f {x_{i - 1} } } = \map f {x_i} - \map f {x_{i - 1} }$

Therefore:

 $\ds \map {V_f} P$ $=$ $\ds \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {\map f {x_i} - \map f {x_{i - 1} } }$ $\ds$ $=$ $\ds \map f {x_n} - \map f {x_0}$ Telescoping Series: Example 2 $\ds$ $=$ $\ds \map f b - \map f a$

for all finite subdivisions $P$.

Note that $\map f b - \map f a$ is independent of the subdivision $P$.

Therefore if $f$ is increasing, it is of bounded variation.

Now consider the case of $f$ decreasing.

$\map f {x_i} \le \map f {x_{i - 1} }$

for all $i \in \N$ with $i \le n$.

Hence, by the definition of the absolute value, we instead have:

$\size {\map f {x_i} - \map f {x_{i - 1} } } = -\paren {\map f {x_i} - \map f {x_{i - 1} } }$

Therefore:

 $\ds \map {V_f} P$ $=$ $\ds \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$ $\ds$ $=$ $\ds -\sum_{i \mathop = 1}^n \paren {\map f {x_i} - \map f {x_{i - 1} } }$ $\ds$ $=$ $\ds -\paren {\map f b - \map f a}$ as in the previous calculation $\ds$ $=$ $\ds \map f a - \map f b$

for all finite subdivisions $P$.

Note again that $\map f a - \map f b$ is independent of the subdivision $P$.

Therefore if $f$ is decreasing, it is of bounded variation.

Hence, if $f$ is either increasing or decreasing, it is of bounded variation.

Hence if $f$ is monotone, it is of bounded variation.

$\blacksquare$