Suppose $f(x)$ is absolutely continuous and hence differentiable almost everywhere. Is it true that there are regions where $f$ is continuously differentiable?
Asked
Active
Viewed 787 times
3 Answers
4
$$ f(x) = \sum_{n=0}^\infty \frac{|\sin(2^n\pi x)|}{4^n} $$ is Lipschitz (and therefore absolutely continuous), but fails to be differentiable at every dyadic rational. In particular there is no interval in which it is differentiable, much less continuously so.
hmakholm left over Monica
- 286,031
-
Thanks! How about if the function is not differentiable only at a finite number of points? – eulerup Jan 31 '16 at 20:08
1
As one sees in the other answer, there are Lipschitz functions where there won't be any intervals at all in which $f$ is differentiable. So you can transition this question to asking instead whether a function can be differentiable everywhere in an interval and fail to be continuously differentiable in any subinterval [it does have to be continuously differentiable at some points however].
But that is another question answered already here: Discontinuous Derivatives
B. S. Thomson
- 7,463
0
Maybe one should mention Rademachers Theorem in this context, if it isn't already known to anyone. That is for $f$ locally Lipschitz it holds that $f$ is differntiable Lebesgue almost everywhere. See e.g. Evans: Measure Theory And Fine Properties Of Functions.
That is of course in no contradiction to the examples already stated as those functions are not differentiable on sets that may be dense but still have Lebesgue measure $0$.
aedolfi
- 1