# Continuously differentiable functions of bounded variation

From this question, we know that a continuous function of bounded variation is not necessarily absolutely continuous. But the example (Devil's staircase) given is not differentiable. What if we require that the function is not only continuous but also continuously differentiable? Is every $C^1$ function that is BV absolutely continuous? Does it matter if we restrict the domain?

Sorry if these questions are too easy. I'm still getting used to these definitions.

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## 2 Answers

This answer actually answers your question, too. (See also point (2) here.)

Yes. The antiderivative of an integrable function is absolutely continuous. If $f$ is $C^1$ and of bounded variation, then $\int \lvert f'\rvert = V(f) < \infty$. So $f$ is the antiderivative of an integrable function.

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Thank you for your help! –  user65431 Mar 16 '13 at 14:27
You are welcome. –  begeistzwerst Mar 16 '13 at 14:27

You don't even need to require bounded variation. Every $C^1$-function is absolutely continuous by the fundamental theorem of calculus: The derivative exists everywhere and integrates back to the original function (even by Riemann integration!).

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The function $x \mapsto x^2$ is not absolutely continuos on an unbounded interval. –  begeistzwerst Mar 14 '13 at 16:50
Fair enough, you are right. I was thinking of a compact interval. But I think if you require bounded variation, then it works even for unbounded domains. –  Thomas Mar 15 '13 at 10:12