1
$\begingroup$

Claim: Let $f : [0,1] \to [0,1]$ be continuous and differentiable almost everywhere on $[0,1]$. If the derivative of $f$ is positive wherever it exists, then $f$ is strictly increasing.

Here's my fallacious proof:

By way of contradiction, suppose there exist $x < y$ in $[0,1]$ such that $f(x) \ge f(y)$. I think I can say by continuity (intermediate value theorem?) that $f(x) \ge f(z)$ whenever $z \in [x,y]$. Now, if for every $z \in (x,y]$ we had that $f$ wasn't differentiable on $(x,z)$, then this would contradict the fact that $f$ is differentiable almost everywhere. Hence, there must exist a $z \in (x,y]$ such that $f$ is differentiable on $(x,z)$. By the mean value theorem, there is a $c \in (x,z)$ such that $f'(c) = \frac{f(z)-f(x)}{z-x} \le 0$, which is a contradiction. Hence, $f$ must be strictly increasing.

As Ryan Unger pointed out in the chatroom, I haven't given a terribly convincing reason why $f$ should be differentiable on any open interval in $[0,1]$, let alone $(x,z)$. So, my question is twofold. First, is the above claim true; is there any way to salvage my proof?

My next question is, does there exist a continuous function which is differentiable almost everywhere but the set of points of differentiability contains no intervals? I was thinking maybe the fat cantor set could help...?

EDIT: I should point out that $f$ doesn't have to have the domain and codomain that I gave it; I only specified those because I'm thinking about Thompson's group $F$.

$\endgroup$
  • $\begingroup$ See Goldowsky-Tonelli theorem. $\endgroup$ – Gabriel Romon Sep 8 at 14:38
  • $\begingroup$ @GabrielRomon Any statement of Goldowsky-Tonelli's theorem I've found says the theorem holds for countable sets. Can it be generalized to sets of measure $0$? $\endgroup$ – user193319 Sep 8 at 15:08
2
$\begingroup$

What about $f(x) = 1-\phi(x)$, where $\phi$ is the Cantor function? $\phi$ maps $[0,1]$ onto $[0,1]$, is continuous and non-decreasing, $\phi'(x)$ exists and is equal to $0$ in the complement of the Cantor set (hence a.e.).

$\endgroup$
  • $\begingroup$ But doesn't the complement of the cantor set contain intervals? $\endgroup$ – user193319 Sep 8 at 16:15
  • $\begingroup$ Yes, it is a countable union of disjoint open intervals. $\endgroup$ – John Dawkins Sep 8 at 16:16
  • 2
    $\begingroup$ Take $f(x)=x-\phi(x)$ instead an you get an actual counterexample, with a derivative that is positive whenever it exists. $\endgroup$ – Henning Makholm Sep 8 at 16:26
  • $\begingroup$ @Henning Makholm: Yes! It's early still and I was misreading "positive"! $\endgroup$ – John Dawkins Sep 8 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.