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 $F$.

  • $\begingroup$ See Goldowsky-Tonelli theorem. $\endgroup$ – Gabriel Romon Sep 8 '19 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 '19 at 15:08

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.).

  • $\begingroup$ But doesn't the complement of the cantor set contain intervals? $\endgroup$ – user193319 Sep 8 '19 at 16:15
  • $\begingroup$ Yes, it is a countable union of disjoint open intervals. $\endgroup$ – John Dawkins Sep 8 '19 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. $\endgroup$ – hmakholm left over Monica Sep 8 '19 at 16:26
  • $\begingroup$ @Henning Makholm: Yes! It's early still and I was misreading "positive"! $\endgroup$ – John Dawkins Sep 8 '19 at 16:29

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