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.
Solution 1:
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 $|f(\beta_n) - f(\alpha_n)| = \Theta(n^{-1/2})$.