Let $f: \mathbb{R} \to \mathbb{R} $ be a differentiable function. Is it true that $f$ is strictly increasing on $\mathbb R$ if and only if $f'(x) \geq 0$ on $\mathbb R$ and the set $X = \{x \in \mathbb{R} \,| \, f'(x) = 0\}$ is countable?
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By increasing do you mean strictly increasing? – Ted Shifrin Jun 05 '20 at 03:17
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i doubt it, since strictly increasing would imply that f'(x) cannot equal 0. – jacob bradley Jun 05 '20 at 03:21
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I suspect OP means monotone increasing. – Dylan C. Beck Jun 05 '20 at 03:25
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@jacobbradley That's not true. $f(x) = x^3$ is strictly increasing but $f'(0) = 0$. I doubt that the OP means nondecreasing, as $f(x) = 0$ would be a trivial counterexample. – Jun 05 '20 at 03:40
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oh yeah ofcourse, my bad. I'd say it would be true if the question was asking about strictly increasing then, since any countable points where f'(x) = 0 would mean that there must be an infinite number of reals in between them – jacob bradley Jun 05 '20 at 03:45
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1I edit it. So it is strictly increasing – 추민서 Jun 05 '20 at 03:48
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This may be of use: https://math.stackexchange.com/a/1999622/169852 – Jun 05 '20 at 03:51
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1I believe you can get the middle thirds Cantor set as the zero set of the derivative of such a function by integration from $0$ to $x$ of the distance to the Cantor set. This gives an example for $f:[0,\infty) \rightarrow {\mathbb R},$ which can easily be extended to a strictly increasing differentiable function (with nonnegative derivative) on all of ${\mathbb R}.$ – Dave L. Renfro Jun 05 '20 at 08:01
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Suppose $f$ is differentiable and strictly increasing. Then yes, we must have $f'\ge 0$ everywhere. However, we can have $f'=0$ on an uncountable set, in fact a set of positive measure.
Example: Let $U$ be an open set containing the rationals, with $m(U)<1.$ Let $E=\mathbb R\setminus U.$ Then $E$ is closed. Define
$$f(x)=\int_0^x d(t,E)\,dt.$$
Since $d(t,E)$ is continuous, $f'(x) = d(x,E)$ everywhere. We thus have $f'=0$ on $E$ and $f'>0$ on $U.$
Now $m(E)=\infty$ and hence is uncountable. The fact that $f'>0$ on the open dense set $U$ implies $f$ is strictly increasing, and we have our example.
zhw.
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if $Q={q_1,q_2,...}$, then $U=\cup _{n=1} ^\infty (q_n - \frac {1} {2^n} , q_n + \frac {1} {2^n})$? – 추민서 Jun 06 '20 at 00:30
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+1, I like this answer better than any of the ones to the question to which this one was duped. – Jun 06 '20 at 20:40
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Edit: the op has edited the question to strictly increasing from simply increasing thus my answer is no longer valid. I think this isn't true and here is why, Take $f(x) = -x^2$ when $x<0$, $f(x) = 0$ when $x \in [0,1]$ and $f(x) = x^2-1$ when $x>1$, this function is clearly continuous, differentiable and increasing (not strictly). Now for the set of points between [0,1] f'(x) = 0, but x is an element of the reals so any closed interval such as [0,1] is uncountable.
jacob bradley
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