12
$\begingroup$

I read from a book without proof the following theorem: Let $f(x)$ be a differentiable function, then $f(x)$ is strictly increasing if and only if $f'(x) \geq 0$, and $f'(x) \gt 0$ almost everywhere. Is it correct? I doubt, but I could neither prove it nor give a counter example.

  • 2
    Monotonic should be replaced by increasing.2011-11-01
  • 1
    Do you agree that $f$ is increasing iff its derivative is non negative? Is it the "almost everywhere positive" part that is posing a problem? The first part has one obvious direction : if you are increasing then the derivative must be everywhere non negative. The other direction (non negative derivative implies increasing) follows from Rolle : if $a2011-11-01
  • 0
    yes, the problem is the "almost everywhere positive" part.2011-11-01
  • 2
    One direction is not too hard : if $f'$ is $\geq 0$ almost everywhere $>0$ then $f$ must be stricly increasing. For take $a$f$ increasing. On the other hand, since $f'>0$ almost everywhere, you can find $a0$ and by definition of $f'(c)$ as a limit you can find $c$c$ with $$\frac{f(d)-f(c)}{d-c}>\frac{1}{2}f'(c)>0.$$ Thus $$f(a)\leq f(c)$f$ is stricly increasing. The two $\leq$ come from $f$ being increasing, while the $<$ comes from the difinition of $c$ and $d$. – 2011-11-01
  • 0
    So actually it's a sufficient additional condition that $f'(x)>0$ on a dense set. This is also necessary, due to the mean value theorem.2011-11-01
  • 0
    @Henning Makholm True. So the actual theorem should read for $f$ differentiable *f stricly increasing iff $f'\geq 0$ and $f'>0$ on a dense set.*.2011-11-01
  • 0
    @Olivier, I'm not convinced that the "almost everywhere" version cannot also be true. Perhaps "$f'>0$ on a dense set" implies "$f'>0$ almost everywhere". If only $f$ were $C^1$ we could probably argue this by continutity, but that isn't necessarily the case. _Or is it?_2011-11-01
  • 0
    The same question is being asked on mathoverflow.2011-11-01
  • 0
    @Henning Makholm I'm wondering the same thing!2011-11-01
  • 0
    @AliBleybel care to share a link to said MO question?2011-11-01
  • 0
    http://mathoverflow.net/questions/79671/sufficient-and-necessary-condition-for-strictly-monotone-differentiable-functions2011-11-01
  • 0
    I'll quote my comment from mathoverflow here also, which gives a counterexample to the statement in the question. "...look at the integral of a continuous function vanishing on the fat cantor set $C$. Eg, $f(x)=\int g(x)\,dx$ with $g(x)$ being the distance from $x$ to $C$ will do. Then $f^\prime=0$ on $C$, which has positive measure, but $f$ is continuously differentiable and strictly increasing."2011-11-01
  • 0
    How can $f'=0$ on $C$ which is dense (I suppose) and still $f'\neq 0$ everywhere? Or maybe I am saying nonsense?2011-11-01
  • 0
    $C$ is not dense. See my answer (and I linked to the Wikipedia page describing $C$).2011-11-01
  • 0
    @Olivier: If $f^\prime=0$ almost everywhere and the derivative of $f$ exists everywhere, then $f$ must be constant. It need not be constant if you don't also require the everywhere differentiability of $f$.2011-11-01
  • 0
    @Henning Makholm: No, my counterexample shows that is false even for $f\in C^1$ (you can even improve my counterexample to be $C^\infty$).2011-11-01
  • 0
    @George Lowther How do you see this? (the fact that $f'=0$ almost everywhere and $f$ everywhere differentiable implies $f$ constant) Also what happens when you only ask for $f'=0$ on a dense set?2011-11-01
  • 0
    @Olivier: Using a bit of measure theory you can write $f(b)-f(a)=\int_a^bf^\prime(x)\,dx$ if $f$ is increasing and everywhere differentiable.2011-11-01
  • 0
    So $f$ would be absolutely continuous? I really don't know much about this stuff ^^2011-11-01
  • 0
    @Olivier: Yes. Everywhere differentiable increasing functions are absolutely continuous. I couldn't give a reference, it's just something I worked out. But if you understand the proof of Lebesgue's theorems concerning differentiability of finite variation functions, it's not too hard to generalize from there.2011-11-01
  • 0
    @George Lowther I will give it a shot tomorow morning!2011-11-01
  • 0
    @Olivier: Just requiring $f^\prime=0$ on a dense set...hmm. With the other conditions, I think that will imply that it is constant, but I can't quite see it yet...2011-11-01
  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1692/discussion-between-george-lowther-and-olivier-begassat)2011-11-01

2 Answers 2

12

A necessary and sufficient condition is that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense subset of $\mathbb{R}$. The condition stated in the question is sufficient, but not necessary. Consider the following differentiable and strictly increasing function for which $f^\prime > 0$ does not hold almost everywhere. Let $C$ be the fat Cantor set (which is closed with positive measure, but does not contain any nontrivial open intervals). Let $g(x)$ be the distance from any point $x$ to $C$. Then, $g$ is continuous and vanishes precisely on $C$. The integral $$ f(x)=\int_0^xg(y)\,dy $$ is strictly increasing and continuously differentiable, but $f^\prime=0$ on $C$.

We can show that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense set is indeed necessary and sufficient for $f$ to be strictly increasing (see also Olivier Bégassat's comments to the question).

Sufficiency: If $f^\prime\ge0$ then the mean value theorem implies that $f$ is increasing. If it was not strictly increasing then we would have $f(a)=f(b)$ for some $a < b$ implying that $f^\prime=0$ on $[a,b]$, contradicting the condition that $f^\prime > 0$ on a dense set. This proves sufficiency of the conditions.

Necessity: In the other direction, suppose that $f$ is strictly increasing. Then, $f^\prime\ge0$ follows directly from the definition of the derivative. Also, $f(a) < f(b)$ for any $a < b$. The mean value theorem implies that $f^\prime(x) > 0$ for some $x\in[a,b]$. So, $f^\prime > 0$ on a dense set.

  • 0
    Your example shows that $f$ stricly increasing need not imply that $f'$ be positve almost everywhere. I was wondering wether $f'=0$ on a dense set implies $f$ constant. All being the same of course : $f$ is still increasing and differentiable everywhere.2011-11-01
  • 0
    @George: Thank you for the detailed answer.2011-11-01
1

As others have said, the proposed "almost everywhere" criterion is not correct: it is sufficient, but not necessary, for $f$ to be strictly increasing.

In his very nice answer, George Lowther has given a correct necessary and sufficient condition: namely that $f'(x) > 0$ on a dense subset of the domain. As it happens, I recently covered this topic in my "Spivak calculus class" and found that even Spivak's text didn't give quite the treatment of this point that I wanted. So I wrote something up myself. Here is the statement:

[Warning: where others here say "strictly increasing", I say "increasing"; where others say "increasing", I say "weakly increasing", and similarly for "monotone" and "weakly monotone".]


Second Monotone Function Theorem: Let $f: I \rightarrow \mathbb{R}$ be a function which is continuous on $I$ and differentiable on the interior $I^{\circ}$ of $I$.
a) The following are equivalent:
(i) $f$ is weakly monotone.
(ii) Either we have $f'(x) \geq 0$ for all $x \in I^{\circ}$ or $f'(x) \leq 0$ for all $x \in I^{\circ}$.
b) Suppose $f$ is weakly monotone. The following are equivalent:
(i) $f$ is not monotone.
(ii) There exist $a,b \in I^{\circ}$ with $a < b$ such that the restriction of $f$ to $[a,b]$ is constant.
(iii) There exist $a,b \in I^{\circ}$ with $a < b$ such that $f'(x) = 0$ for all $x \in [a,b]$.


Note that my statement of the equivalence is slightly different (and, perhaps, slightly simpler?) than George's. For the proof, please see $\S 5.1$ of these notes. Finally, note that when George appeals to the Intermediate Value Theorem I think he really means to appeal (as I do) to the Mean Value Theorem, although (confusingly, for students) these names are not completely standardized even among mathematicians.

  • 0
    I did intend to use the mean value theorem. I corrected my answer, thanks!2011-11-01