0
$\begingroup$

This question already has an answer here:

We have "$f'(x)>0$ in interval $I\implies f(x)$ strictly increasing in interval $I$".
However, the converse is not true.
A frequently cited example is the function $f(x)=x^3$, which is strictly increasing but $f'(0)=0$.
Here comes my question:

What is the necessary and sufficient condition of $f'(x)$ to make $f(x)$ strictly increasing?

My guess is:

$f'(x)>0$ almost everywhere.

This is clearly a sufficient condition, as we can use Lebesgue integration and "drop the word 'almost'".

But is this a necessary condition? Is there a weaker condition? A full solution or a little hint will both be appreciated. Thank you in advance.

$\endgroup$

marked as duplicate by Paramanand Singh, Leucippus, Joey Zou, R_D, Claude Leibovici calculus Oct 6 '16 at 5:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If $f $ is absolutely continuous then it's true by Lebesgue differentiation. Otherwise it's false and the Cantor function gives an example of a strictly increasing function that has zero derivative a.e. $\endgroup$ – Jose27 Oct 5 '16 at 17:34
  • $\begingroup$ See Constructing a strictly increasing function with zero derivatives. $\endgroup$ – dxiv Oct 5 '16 at 17:36
  • $\begingroup$ Can we assume that $f(x)$ is differentiable on $\mathbb R$ , otherwise strictly increasing functions not even continous are possible. $\endgroup$ – Peter Oct 5 '16 at 17:43
1
$\begingroup$

Assume $f$ is differentiable on an interval $I$ and $f'(x)\ge 0$ on $I.$ Let $Z = \{x \in I: f'(x)=0\}.$ Then $f$ is strictly increasing on $I$ iff $Z$ contains no interval. (Here "interval" means "interval of positive length".)

Proof: Suppose $f$ is strictly increasing on $I.$ If $Z$ contained an interval $J,$ then $f'\equiv 0$ on $J,$ which implies $f$ is constant on $J$ by the mean value theorem. Thus $f$ is not strictly increasing, contradiction.

Suppose $Z$ contains no interval. Let $x,y\in I$ with $x<y.$ Because $f'\ge 0,$ we have $f(x)\le f(z) \le f(y)$ for all $z\in [x,y].$ If $f(y) = f(x),$ then $f = f(x)$ on $[x,y].$ This implies $f'\equiv 0$ on $[x,y],$ hence $Z$ contains $[x,y],$ contradiction. The contradiction shows $f(x) < f(y),$ hence $f$ is strictly increasing.

$\endgroup$
0
$\begingroup$

If $f(x)$ satisfies $f'(x)\ge 0$ for all $x\in \mathbb R$ and if for every $x_0\in \mathbb R$ with $f'(x_0)=0$, there is a neighborhood around $x_0$ , such that $f'(x)\ne 0$ within this neighborhood (except $x_0$) (in other words : isolated roots of the derivate ) , then $f(x)$ is strictly increasing on $\mathbb R$.

Not sure, whether we have a weaker sufficient condition, but an interval $[a,b]$ with $a<b$ and $f'(x)=0$ for all $x\in [a,b]$ is obviously impossible.

$\endgroup$

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