Tell me more ×  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.  share|improve this question 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 add comment  1 Answer  up vote 4 down vote accepted
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
$|f(\alpha_n) - f(\beta_n)| = \Theta(n^{-1/2})$.  share|improve
this answer 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)