# Does the everywhere differentiability 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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No. Consider $f(x) = x^2 \sin(1/x^4)$ on $[-1,1]$. Note that for $\beta_n = (\pi n)^{-1/4}$ and $\alpha_n = (\pi (n+1/2))^{-1/4}$ we have $\beta_n - \alpha_n = \Theta( n^{-5/4})$ while $|f(\beta_n) - f(\alpha_n)| = \Theta(n^{-1/2})$.

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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
@Robert Israel: How doe we conclude from the above that $f$ is not absolutely continuous ? Sorry if I am missing some obvious step after your explanation. – pikachuchameleon Jun 19 '15 at 18:27
Definition of absolute continuity: for any $\epsilon > 0$ there exists $\delta > 0$ such that for any finite sequence of intervals $[\alpha_n, \beta_n]$ with $\sum_j (\beta_j - \alpha_j) < \delta$, $\sum_j |f(\beta_j) - f(\alpha_j)| < \epsilon$. In this case, since $\sum_n n^{-5/4}$ converges but $\sum_n n^{-1/2}$ diverges, for any $\epsilon$ and $\delta$ you can take $(\alpha_n, \beta_n)$ for $M \le n < N$ that gives you $\sum_j (\beta_j - \alpha_j) < \delta$ but $\sum_j |f(\beta_j) - f(\alpha_j)| > \epsilon$. – Robert Israel Jun 19 '15 at 19:13
What is $f(0)$? – Jack Mar 5 at 15:18