How to prove that x^1/3 is absolutely continuous without using

the fact that it is the integral of a derivative.

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- Dec 3rd 2011, 04:30 PMhgd7833Absolute continuity
How to prove that x^1/3 is absolutely continuous without using

the fact that it is the integral of a derivative. - Dec 3rd 2011, 05:36 PMDrexel28Re: Absolute continuity
I remember this being annoying to do. You know everything's fine on any union of intervals bounded away from zero, since $\displaystyle x^{\frac{1}{3}}$ has bounded derivative, and thus is Lipschitz. Try then to take care of the case of finite intervals containing $\displaystyle 0$. Is this where you're stuck?

- Dec 4th 2011, 07:58 AMhgd7833Re: Absolute continuity
I would like to prove that f is absolute continuous over the closed interval [-1,1] .

Actually the derivative is unbounded there thus not Lipschitz.

I am trying to use the definition.