# Does the everywhere differentiablity of $f$ imply it is absolutely continuous on a compact interval?

Suppose $f$ is differentiable everywhere on $[0,1]$. Must $f$ be absolutely continuous on $[0,1]$? I know this is true if $f'$ is integrable but I'm not sure in this more general case.

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If a function is absolutely continuous, then it's differentiable almost every where. The converse is not true in general. See the equivalent definitions of absolute continuity. –  Mhenni Benghorbal Aug 23 '12 at 22:09

No. Consider $f(x) = x^2 \sin(1/x^4)$ on $[-1,1]$. Note that for $\alpha_n = (\pi n)^{-1/4}$ and $\beta_n = (\pi (n+1/2))^{-1/4}$ we have $\alpha_n - \beta_n = \Theta( n^{-5/4})$ while $|f(\alpha_n) - f(\beta_n)| = \Theta(n^{-1/2})$.
It's worth mentioning that there's a theorem that says that if $f$ is absolutely continuous, then $f'$ is integrable, so you can also show this example works by looking at the formula for $f'$ and seeing it blows up like ${1 \over x^3}$ (in a sinusoidal fashion) as $x$ goes to zero. –  Zarrax Aug 22 '12 at 5:47