Differentiable Function of Bounded Variation may not have Bounded Derivative

Theorem

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

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

Let $f$ be differentiable on $\openint a b$.

Then $f'$ is not necessarily bounded.

Proof

Proof by Counterexample:

Take $a = 0$, $b = 1$.

Let $f : \closedint 0 1 \to \R$ have:

$\map f x = \sqrt x$

for all $x \in \closedint 0 1$.

Note that $f$ is increasing, so by Monotone Function is of Bounded Variation:

$f$ is of bounded variation.

By Derivative of Power, $f$ is differentiable on $\openint 0 1$ with derivative:

$\map {f'} x = \dfrac 1 {2 \sqrt x}$

Note that this is unbounded as $x \to 0^+$.

So $f$ does not have a bounded derivative, despite being of bounded variation.

$\blacksquare$

Also see

• Differentiable Function with Bounded Derivative is of Bounded Variation, of which this is the (false) converse.