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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How to prove that x^1/3 is absolutely continuous without using
the fact that it is the integral of a derivative.
I remember this being annoying to do. You know everything's fine on any union of intervals bounded away from zero, sinceQuote:
Originally Posted by hgd7833
has bounded derivative, and thus is Lipschitz. Try then to take care of the case of finite intervals containing
. Is this where you're stuck?
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.