Differentiable Function with Bounded Derivative 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 continuous function.

Let $f$ be differentiable on $\openint a b$, with bounded derivative.


Then $f$ is of bounded variation.


Proof

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

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

with:

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

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

$\size {\map {f'} x} \le M$

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

By the Mean Value Theorem, for each $i \in \N$ with $i \le n$, there exists $\xi_i \in \openint {x_{i - 1} } {x_i}$ such that:

$\map {f'} {\xi_i} = \dfrac {\map f {x_i} - \map f {x_{i - 1} } } {x_i - x_{i - 1} }$

Note that, from the boundedness of $f'$:

$\size {\map {f'} {\xi_i} } \le M$

We also have from the fact that $x_i > x_{i - 1}$:

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

So, for each $i$:

\(\ds \size {\map f {x_i} - \map f {x_{i - 1} } }\) \(=\) \(\ds \size {\map {f'} {\xi_i} } \paren {x_i - x_{i - 1} }\)
\(\ds \) \(\le\) \(\ds M \paren {x_i - x_{i - 1} }\)

We therefore have:

\(\ds \map {V_f} {P ; \closedint a b}\) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }\) using the notation from the definition of bounded variation
\(\ds \) \(\le\) \(\ds M \sum_{i \mathop = 1}^n \paren {x_i - x_{i - 1} }\)
\(\ds \) \(=\) \(\ds M \paren {x_n - x_0}\) Telescoping Series: Example 2
\(\ds \) \(=\) \(\ds M \paren {b - a}\)

for all finite subdivisions $P$.

So $f$ is of bounded variation.

$\blacksquare$


Also see


Sources