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Let $f(x)$ be a Lipschitz real-valued function defined on a closed interval $I$. The derivative $f '(x)$ exists a.e. since $f$ is absolutely continuous.

My question is: Is $f '(x)$ necessarily square integrable, i.e. in $L^2(I)$ ?

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up vote 7 down vote accepted

Yes, because the derivative is bounded. If $M$ is the Lipschitz constant, then $|f'(x)|\leq M$ everywhere it exists. This follows from the fact that each of the difference quotients has this bound. Lipschitz continuous functions on $I$ are precisely the indefinite integrals of bounded measurable functions on $I$, whereas AC functions are the indefinite integrals of integrable functions on $I$. (Thus you find AC counterexamples to this by integrating integrable but not square integrable functions.)

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