Let $f$ be a real-valued, everywhere differentiable function on $[0,1]$. Suppose $f \in BV[0,1]$. Prove $f \in AC[0,1]$.

Asked Jul 16 '23 at 19:37

Active Jul 17 '23 at 00:06

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I tried to prove this by showing that the derivative $f'$ must be bounded. But I am not sure this is true, as the Extreme Value Theorem cannot be invoked ($f'$ is not guaranteed to be continuous) nor is is the case that being of bounded variation implies that the derivative is bounded (this is true the other way around).

The following proof here (Are differentiable functions of bounded variation on [0,1] also absolutely continuous?) technically works, but it cites what seems to be an obscure theorem from Rudin (7.1), which I am not sure is allowed to be assumed in the proof to this question. This was a qualifying exam question and only "well-known" theorems are allowed to be cited.

edited Jul 17 '23 at 00:06

asked Jul 16 '23 at 19:37

kam

Indeed, there are differentiable functions of bounded variation for which the derivative isn't bounded, such an example can be found here: Does differentiable function of bounded variation have bounded derivative? – Bruno B Jul 16 '23 at 19:47
@BrunoB Not quite, as I'm not sure if that theorem 7.21 from Rudin referred in your answer is allowed to be assumed in the proof. – kam Jul 16 '23 at 23:32
It is not an obscured theorem. It is discussed in many books on integration. Is a result due to Walker – Mittens Jul 17 '23 at 04:18

Let $f$ be a real-valued, everywhere differentiable function on $[0,1]$. Suppose $f \in BV[0,1]$. Prove $f \in AC[0,1]$.

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