Differentiable Function with Bounded Derivative is of Bounded Variation

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 $i$:

We therefore have:

for all finite subdivisions $P$.

So $f$ is of bounded variation.

Also see

 * Differentiable Function of Bounded Variation may not have Bounded Derivative: demonstrating that while this theorem gives a simple sufficient condition for a differentiable continuous function $f$ to be of bounded variation, it is not a necessary condition.