I am trying to understand absolute continuity and continuous nowhere-differentiable functions, and since are more than one criteria for continuity I am a bit lost.
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It is the function $q(x) = \frac{x}{2}\log(x^2)$ absolute continuous near zero? Since it "softly" reaches an infinite slope at zero ($q(x)$ is continuous and also $q'(x)$ except at $x=0$ which is a measure-zero point), I am not sure if it is fulfilling absolute convergence.
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It is the function $s(x) = |\frac{x}{2}\log(x^2)|$ absolute continuous near zero? Now is continuous but surely not differentiable at $x=0$ (has a sharp point like the absolute value function). I want to know if these change things about the absolute convergence compared with the previous function.
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It is the Weierstrass function fulfilling absolute convergence?
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It is the Blancmange curve fulfilling absolute convergence?
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It is the Wiener process fulfilling absolute convergence? Here I think that the answer is negative, since Wiener Process is of unbounded variation, and even when the Cantor function is of bounded variation when not fulfilling absolute continuity, from Wikipedia it looks like every absolute continuous function must be of bounded variation, so Wiener process wouldn´t be fulfilling that. I don´t know if the same argument could be extended to the previous continuous nowhere-differentiable functions.
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There are examples of continuous nowhere-differentiable functions that are absolute continuous?
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It is absolute continuity incompatible with a function being continuous but nowhere-differentiable?
Best Answer
An absolutely continuous function is differentiable almost everywhere and is equal to the integral of its derivative. This is one way to characterize absolutely continuous functions. The equivalence with the standard definition is due to Lebesgue. It is connected to the Lebesgue differentiation theorem.
Absolutely continuous functions on an interval $[a,b]$ may be characterized as distributions whose derivatives are integrable functions.
For $ {x \over 2} \log(x^2) = x\log(x)$, $${d\over dx} (x\log(x)) = 1 + \log(x)$$ This is integrable near the origin hence we get absolute continuity.