Monotone Function is of Bounded Variation

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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.

We instead have:

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

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.

$\blacksquare$


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